JavaScript 面试中常见算法问题详解

数字

20# Leetcode 378. Kth Smallest Element in a Sorted Matrix

解释下 Prototypal Inheritance 与 Classical Inheritance 的区别

在类继承中,类是不可变的,不同的语言中对于多继承的支持也不一样,有些语言中还支持接口、final、abstract
的概念。而原型继承则更为灵活,原型本身是可以可变的,并且对象可能继承自多个原型。

40# LintCode: Maximum Number in Mountain Sequence

Given a mountain sequence of n integers which increase firstly and then
decrease, find the mountain top.

Example
Given nums = [1, 2, 4, 8, 6, 3] return 8
Given nums = [10, 9, 8, 7], return 10

public int mountainSequence(int[] nums) {
    if (nums.length == 0 || nums == null) {
        return -1;
    }
    int start = 0;
    int end = nums.length - 1;
    while (start + 1 < end) {
        int mid = start + (end - start) / 2;
        if (nums[mid] > nums[mid + 1]) {
            end = mid;
        } else {
            start = mid;
        }
    }
    return Math.max(nums[start], nums[end]);
}

颠倒字符串

给定某个字符串,要求将其中单词倒转之后然后输出,譬如”Welcome to this
Javascript Guide!” 应该输出为 “emocleW ot siht tpircsavaJ !ediuG”。

JavaScript

var string = “Welcome to this Javascript Guide!”; // Output becomes
!ediuG tpircsavaJ siht ot emocleW var reverseEntireSentence =
reverseBySeparator(string, “”); // Output becomes emocleW ot siht
tpircsavaJ !ediuG var reverseEachWord =
reverseBySeparator(reverseEntireSentence, ” “); function
reverseBySeparator(string, separator) { return
string.split(separator).reverse().join(separator); }

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var string = "Welcome to this Javascript Guide!";
 
// Output becomes !ediuG tpircsavaJ siht ot emocleW
var reverseEntireSentence = reverseBySeparator(string, "");
 
// Output becomes emocleW ot siht tpircsavaJ !ediuG
var reverseEachWord = reverseBySeparator(reverseEntireSentence, " ");
 
function reverseBySeparator(string, separator) {
  return string.split(separator).reverse().join(separator);
}

32# Leetcode 4. Median of Two Sorted Arrays

There are two sorted arrays nums1 and nums2 of size m and n
respectively.

Find the median of the two sorted arrays. The overall run time
complexity should be O(log (m+n)).

Example 1:
nums1 = [1, 3], nums2 = [2]
The median is 2.0

Example 2:
nums1 = [1, 2], nums2 = [3, 4]
The median is (2 + 3)/2 = 2.5

数组

1# Leetcode 367. Valid Perfect Square

Given a positive integer num, write a function which returns True if
num is a perfect square else False.

Note: Do not use any built-in library function such as sqrt.

Example 1:
Input: 16
Returns: True

Example 2:
Input: 14
Returns: False

思路:
以256举例
mid = 128 => 128 * 128 > 256 => end = mid = 128;
mid = 64 => 64 * 64 > 256 => end = mid = 64;
mid = 32 => 32 * 32 > 256 => end = mid = 32;
mid = 16 => 16 * 16 = 256 => return true;

以15举例
mid = 8 => 8 * 8 > 15 => end = mid = 8;
mid = 4 => 4 * 4 > 15 => end = mid = 4;
mid = 2 => 2 * 2 < 15 => start = mid = 2; end = 4;
mid = 3 => 3 * 3 < 15 => start = mid = 3; end = 4;
start + 1 = 3 + 1 = 4 = end, while loop end;
start = 3, 3 * 3 != 15 and end = 4, 4 * 4 != 15;
so return false;

public class Solution {
    public boolean isPerfectSquare(int num) {
        if (num < 1) {
            return false;
        }
        long start = 1;
        long end = num;
        while (start + 1 < end) {
            long mid = start + (end - start) / 2;
            if (mid * mid == num) {
                return true;
            } else if (mid * mid < num) {
                start = mid;
            } else {
                end = mid;
            }
        }
        if (start * start == num || end * end == num) {
            return true;
        }
        return false;
    }
}

字符串

8# Leetcode 278. First Bad Version

You are a product manager and currently leading a team to develop a
new product. Unfortunately, the latest version of your product fails
the quality check. Since each version is developed based on the
previous version, all the versions after a bad version are also bad.

Suppose you have n versions [1, 2, …, n] and you want to find out
the first bad one, which causes all the following ones to be bad.

You are given an API bool isBadVersion(version) which will return
whether version is bad. Implement a function to find the first bad
version. You should minimize the number of calls to the API.

/* The isBadVersion API is defined in the parent class VersionControl.
      boolean isBadVersion(int version); */

public class Solution extends VersionControl {
    public int firstBadVersion(int n) {
        int start = 1;
        int end = n;
        while (start + 1 < end) {
            int mid = start + (end - start) / 2;
            if (isBadVersion(mid)) {
                end = mid;
            }
            else {
                start = mid;
            }
        }
        if(isBadVersion(start)) {
            return start;
        }
        return end;
    }
}

数组中元素最大差值计算

给定某无序数组,求取任意两个元素之间的最大差值,注意,这里要求差值计算中较小的元素下标必须小于较大元素的下标。譬如[7, 8, 4, 9, 9, 15, 3, 1, 10]这个数组的计算值是
11( 15 – 4 ) 而不是 14(15 – 1),因为 15 的下标小于 1。

JavaScript

var array = [7, 8, 4, 9, 9, 15, 3, 1, 10]; // [7, 8, 4, 9, 9, 15, 3,
1, 10] would return `11` based on the difference between `4` and
`15` // Notice: It is not `14` from the difference between `15`
and `1` because 15 comes before 1. findLargestDifference(array);
function findLargestDifference(array) { //
如果数组仅有一个元素,则直接返回 -1 if (array.length <= 1) return -1;
// current_min 指向当前的最小值 var current_min = array[0]; var
current_max_difference = 0; //
遍历整个数组以求取当前最大差值,如果发现某个最大差值,则将新的值覆盖
current_max_difference // 同时也会追踪当前数组中的最小值,从而保证
`largest value in future` – `smallest value before it` for (var i =
1; i < array.length; i++) { if (array[i] > current_min &&
(array[i] – current_min > current_max_difference)) {
current_max_difference = array[i] – current_min; } else if
(array[i] <= current_min) { current_min = array[i]; } } // If
negative or 0, there is no largest difference if
(current_max_difference <= 0) return -1; return
current_max_difference; }

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var array = [7, 8, 4, 9, 9, 15, 3, 1, 10];
// [7, 8, 4, 9, 9, 15, 3, 1, 10] would return `11` based on the difference between `4` and `15`
// Notice: It is not `14` from the difference between `15` and `1` because 15 comes before 1.
 
findLargestDifference(array);
 
function findLargestDifference(array) {
 
  // 如果数组仅有一个元素,则直接返回 -1
 
  if (array.length <= 1) return -1;
 
  // current_min 指向当前的最小值
 
  var current_min = array[0];
  var current_max_difference = 0;
  
  // 遍历整个数组以求取当前最大差值,如果发现某个最大差值,则将新的值覆盖 current_max_difference
  // 同时也会追踪当前数组中的最小值,从而保证 `largest value in future` – `smallest value before it`
 
  for (var i = 1; i < array.length; i++) {
    if (array[i] > current_min && (array[i] – current_min > current_max_difference)) {
      current_max_difference = array[i] – current_min;
    } else if (array[i] <= current_min) {
      current_min = array[i];
    }
  }
 
  // If negative or 0, there is no largest difference
  if (current_max_difference <= 0) return -1;
 
  return current_max_difference;
}

11# Leetcode 350. Intersection of Two Arrays II

Given two arrays, write a function to compute their intersection.

Example:
Given nums1 = [1, 2, 2, 1], nums2 = [2, 2], return [2, 2].

Note:
Each element in the result should appear as many times as it shows in
both arrays.
The result can be in any order.
Follow up:
What if the given array is already sorted? How would you optimize your
algorithm?
What if nums1’s size is small compared to nums2’s size? Which
algorithm is better?
What if elements of nums2 are stored on disk, and the memory is
limited such that you cannot load all elements into the memory at
once?

public class Solution {
    public int[] intersect(int[] nums1, int[] nums2) {
        Arrays.sort(nums1);
        Arrays.sort(nums2);
        int index1 = 0;
        int index2 = 0;
        List<Integer> list = new ArrayList<>();
        while(index1 < nums1.length && index2 < nums2.length) {
            if (nums1[index1] == nums2[index2]) {
                list.add(nums1[index1]);
                index1++;
                index2++;
            } else if (nums1[index1] < nums2[index2]) {
                index1++;
            } else if (nums1[index1] > nums2[index2]) {
                index2++;
            }
        }
        int[] result = new int[list.size()];
        int index = 0;
        for (int element: list) {
            result[index++] = element;
        }
        return result;
    }
}

阐述下 use strict; 的作用

use strict; 顾名思义也就是 JavaScript
会在所谓严格模式下执行,其一个主要的优势在于能够强制开发者避免使用未声明的变量。对于老版本的浏览器或者执行引擎则会自动忽略该指令。

JavaScript

// Example of strict mode “use strict”; catchThemAll(); function
catchThemAll() { x = 3.14; // Error will be thrown return x * x; }

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// Example of strict mode
"use strict";
 
catchThemAll();
function catchThemAll() {
  x = 3.14; // Error will be thrown
  return x * x;
}

14# Leetcode 33. Search in Rotated Sorted Array

Suppose an array sorted in ascending order is rotated at some pivot
unknown to you beforehand.

(i.e., 0 1 2 4 5 6 7 might become 4 5 6 7 0 1 2).

You are given a target value to search. If found in the array return
its index, otherwise return -1.

You may assume no duplicate exists in the array.

public class Solution {
    public int search(int[] nums, int target) {
        if (nums == null || nums.length == 0) {
            return -1;
        }
        int start = 0;
        int end = nums.length - 1;
        while (start + 1 < end) {
            int mid = start + (end - start) / 2;
            if (nums[mid] == target) {
                return mid;
            }
            if (nums[start] < nums[mid]) {
                if (nums[start] <= target && target <= nums[mid]) {
                    end = mid;
                } else {
                    start = mid;
                } 
            }
            else {
                if (nums[mid] <= target && target <= nums[end]) {
                    start = mid;
                }
                else {
                    end = mid;
                }
            }
        }
        if (nums[start] == target) {
            return start;
        }
        if (nums[end] ==  target) {
            return end;
        }
        return -1;
    }
}

JavaScript 面试中常见算法问题详解

2017/02/20 · JavaScript
· 1 评论 ·
算法

原文出处:
王下邀月熊_Chevalier   

JavaScript
面试中常见算法问题详解
翻译自
Interview Algorithm Questions in Javascript()
{…}

从属于笔者的 Web
前端入门与工程实践
。下文提到的很多问题从算法角度并不一定要么困难,不过用
JavaScript 内置的 API 来完成还是需要一番考量的。

30# Leetcode 209. Minimum Size Subarray Sum

判断大括号是否闭合

创建一个函数来判断给定的表达式中的大括号是否闭合:

JavaScript

var expression = “{{}}{}{}” var expressionFalse = “{}{{}”;
isBalanced(expression); // true isBalanced(expressionFalse); // false
isBalanced(“”); // true function isBalanced(expression) { var
checkString = expression; var stack = []; // If empty, parentheses are
technically balanced if (checkString.length <= 0) return true; for
(var i = 0; i < checkString.length; i++) { if(checkString[i] ===
‘{‘) { stack.push(checkString[i]); } else if (checkString[i] ===
‘}’) { // Pop on an empty array is undefined if (stack.length > 0) {
stack.pop(); } else { return false; } } } // If the array is not empty,
it is not balanced if (stack.pop()) return false; return true; }

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var expression = "{{}}{}{}"
var expressionFalse = "{}{{}";
 
isBalanced(expression); // true
isBalanced(expressionFalse); // false
isBalanced(""); // true
 
function isBalanced(expression) {
  var checkString = expression;
  var stack = [];
 
  // If empty, parentheses are technically balanced
  if (checkString.length <= 0) return true;
 
  for (var i = 0; i < checkString.length; i++) {
    if(checkString[i] === ‘{‘) {
      stack.push(checkString[i]);
    } else if (checkString[i] === ‘}’) {
      // Pop on an empty array is undefined
      if (stack.length > 0) {
        stack.pop();
      } else {
        return false;
      }
    }
  }
 
  // If the array is not empty, it is not balanced
  if (stack.pop()) return false;
  return true;
}

34# Leetcode 354. Russian Doll Envelopes

== 与 === 的区别是什么

=== 也就是所谓的严格比较,关键的区别在于===
会同时比较类型与值,而不是仅比较值。

JavaScript

// Example of comparators 0 == false; // true 0 === false; // false 2 ==
‘2’; // true 2 === ‘2’; // false

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// Example of comparators
0 == false; // true
0 === false; // false
 
2 == ‘2’; // true
2 === ‘2’; // false

36# Leetcode 302.Smallest Rectangle Enclosing Black Pixels

阐述下 JavaScript 中的变量提升

所谓提升,顾名思义即是 JavaScript
会将所有的声明提升到当前作用域的顶部。这也就意味着我们可以在某个变量声明前就使用该变量,不过虽然
JavaScript 会将声明提升到顶部,但是并不会执行真的初始化过程。

27# Leetcode 50. Pow(x, n)

寻找连续数组中的缺失数

给定某无序数组,其包含了 n 个连续数字中的 n – 1
个,已知上下边界,要求以O(n)的复杂度找出缺失的数字。

JavaScript

// The output of the function should be 8 var array_of_integers = [2,
5, 1, 4, 9, 6, 3, 7]; var upper_bound = 9; var lower_bound = 1;
findMissingNumber(array_of_integers, upper_bound, lower_bound); //8
function findMissingNumber(array_of_integers, upper_bound,
lower_bound) { // Iterate through array to find the sum of the numbers
var sum_of_integers = 0; for (var i = 0; i <
array_of_integers.length; i++) { sum_of_integers +=
array_of_integers[i]; } // 以高斯求和公式计算理论上的数组和 //
Formula: [(N * (N + 1)) / 2] – [(M * (M – 1)) / 2]; // N is the
upper bound and M is the lower bound upper_limit_sum = (upper_bound
* (upper_bound + 1)) / 2; lower_limit_sum = (lower_bound *
(lower_bound – 1)) / 2; theoretical_sum = upper_limit_sum –
lower_limit_sum; // return (theoretical_sum – sum_of_integers) }

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// The output of the function should be 8
var array_of_integers = [2, 5, 1, 4, 9, 6, 3, 7];
var upper_bound = 9;
var lower_bound = 1;
 
findMissingNumber(array_of_integers, upper_bound, lower_bound); //8
 
function findMissingNumber(array_of_integers, upper_bound, lower_bound) {
 
  // Iterate through array to find the sum of the numbers
  var sum_of_integers = 0;
  for (var i = 0; i < array_of_integers.length; i++) {
    sum_of_integers += array_of_integers[i];
  }
 
  // 以高斯求和公式计算理论上的数组和
  // Formula: [(N * (N + 1)) / 2] – [(M * (M – 1)) / 2];
  // N is the upper bound and M is the lower bound
 
  upper_limit_sum = (upper_bound * (upper_bound + 1)) / 2;
  lower_limit_sum = (lower_bound * (lower_bound – 1)) / 2;
 
  theoretical_sum = upper_limit_sum – lower_limit_sum;
 
  //
  return (theoretical_sum – sum_of_integers)
}

9# Leetcode 475. Heaters

Winter is coming! Your first job during the contest is to design a
standard heater with fixed warm radius to warm all the houses.

Now, you are given positions of houses and heaters on a horizontal
line, find out the minimum radius of heaters so that all houses could
be covered by those heaters.

So, your input will be the positions of houses and heaters separately,
and your expected output will be the minimum radius standard of
heaters.

Note:
Numbers of houses and heaters you are given are non-negative and will
not exceed 25000.

Positions of houses and heaters you are given are non-negative and
will not exceed 10^9.

As long as a house is in the heaters’ warm radius range, it can be
warmed.

All the heaters follow your radius standard and the warm radius will
the same.

Example 1:
Input: [1,2,3],[2]
Output: 1
Explanation: The only heater was placed in the position 2, and if we
use the radius 1 standard, then all the houses can be warmed.

Example 2:
Input: [1,2,3,4],[1,4]
Output: 1
Explanation: The two heater was placed in the position 1 and 4. We
need to use radius 1 standard, then all the houses can be warmed.

升序排列加热器的坐标heaters
遍历房屋houses,记当前房屋坐标为house:
利用二分查找,分别找到不大于house的最大加热器坐标left,以及不小于house的最小加热器坐标right(即左右最近的heater),
则当前房屋所需的最小加热器半径radius = min(house – left, right –
house)。利用radius更新最终答案。

public class Solution {
    public int findRadius(int[] houses, int[] heaters) {
        //sort
        Arrays.sort(houses);
        Arrays.sort(heaters);

        int radius = 0;
        for( int house: houses) {
            int local = binarySearch(heaters, house);
            radius = Math.max(radius, local);
        }
        return radius;
    }

    private int binarySearch(int[] heaters, int target) {
        int start = 0;
        int end = heaters.length - 1;
        while (start + 1 < end) {
            int mid = start + (end - start) / 2;
            if (heaters[mid] == target) {
                return 0;
            } else if (heaters[mid] < target) {
                start = mid;
            } else {
                end = mid;
            }
        }
        return Math.min (Math.abs(target - heaters[start]),
                        Math.abs(target - heaters[end]));
    }
}

栈与队列

38# Leetcode 483. Smallest Good Base

判断是否为 2 的指数值

JavaScript

isPowerOfTwo(4); // true isPowerOfTwo(64); // true isPowerOfTwo(1); //
true isPowerOfTwo(0); // false isPowerOfTwo(-1); // false // For the
non-zero case: function isPowerOfTwo(number) { // `&` uses the bitwise
n. // In the case of number = 4; the expression would be identical to:
// `return (4 & 3 === 0)` // In bitwise, 4 is 100, and 3 is 011. Using
&, if two values at the same // spot is 1, then result is 1, else 0. In
this case, it would return 000, // and thus, 4 satisfies are expression.
// In turn, if the expression is `return (5 & 4 === 0)`, it would be
false // since it returns 101 & 100 = 100 (NOT === 0) return number &
(number – 1) === 0; } // For zero-case: function
isPowerOfTwoZeroCase(number) { return (number !== 0) && ((number &
(number – 1)) === 0); }

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isPowerOfTwo(4); // true
isPowerOfTwo(64); // true
isPowerOfTwo(1); // true
isPowerOfTwo(0); // false
isPowerOfTwo(-1); // false
 
// For the non-zero case:
function isPowerOfTwo(number) {
  // `&` uses the bitwise n.
  // In the case of number = 4; the expression would be identical to:
  // `return (4 & 3 === 0)`
  // In bitwise, 4 is 100, and 3 is 011. Using &, if two values at the same
  // spot is 1, then result is 1, else 0. In this case, it would return 000,
  // and thus, 4 satisfies are expression.
  // In turn, if the expression is `return (5 & 4 === 0)`, it would be false
  // since it returns 101 & 100 = 100 (NOT === 0)
 
  return number & (number – 1) === 0;
}
 
// For zero-case:
function isPowerOfTwoZeroCase(number) {
  return (number !== 0) && ((number & (number – 1)) === 0);
}

 

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图片 1

23# Leetcode 436. Find Right Interval

Given a set of intervals, for each of the interval i, check if there
exists an interval j whose start point is bigger than or equal to the
end point of the interval i, which can be called that j is on the
“right” of i.

For any interval i, you need to store the minimum interval j’s index,
which means that the interval j has the minimum start point to build
the “right” relationship for interval i. If the interval j doesn’t
exist, store -1 for the interval i. Finally, you need output the
stored value of each interval as an array.

Note:
You may assume the interval’s end point is always bigger than its
start point.
You may assume none of these intervals have the same start point.
Example 1:
Input: [ [1,2] ]
Output: [-1]
Explanation: There is only one interval in the collection, so it
outputs -1.

Example 2:
Input: [ [3,4], [2,3], [1,2] ]
Output: [-1, 0, 1]
Explanation: There is no satisfied “right” interval for [3,4].
For [2,3], the interval [3,4] has minimum-“right” start point;
For [1,2], the interval [2,3] has minimum-“right” start point.

Example 3:
Input: [ [1,4], [2,3], [3,4] ]
Output: [-1, 2, -1]
Explanation: There is no satisfied “right” interval for [1,4] and
[3,4].
For [2,3], the interval [3,4] has minimum-“right” start point.

二分搜索

JavaScript

function recursiveBinarySearch(array, value, leftPosition,
rightPosition) { // Value DNE if (leftPosition > rightPosition)
return -1; var middlePivot = Math.floor((leftPosition + rightPosition) /
2); if (array[middlePivot] === value) { return middlePivot; } else if
(array[middlePivot] > value) { return recursiveBinarySearch(array,
value, leftPosition, middlePivot – 1); } else { return
recursiveBinarySearch(array, value, middlePivot + 1, rightPosition); } }

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function recursiveBinarySearch(array, value, leftPosition, rightPosition) {
  // Value DNE
  if (leftPosition > rightPosition) return -1;
 
  var middlePivot = Math.floor((leftPosition + rightPosition) / 2);
  if (array[middlePivot] === value) {
    return middlePivot;
  } else if (array[middlePivot] > value) {
    return recursiveBinarySearch(array, value, leftPosition, middlePivot – 1);
  } else {
    return recursiveBinarySearch(array, value, middlePivot + 1, rightPosition);
  }
}

12# Leetcode 153. Find Minimum in Rotated Sorted Array

Suppose an array sorted in ascending order is rotated at some pivot
unknown to you beforehand. (i.e., 0 1 2 4 5 6 7 might become 4 5 6 7 0
1 2).

Find the minimum element.

You may assume no duplicate exists in the array.

public class Solution {
    public int findMin(int[] nums) {
        if (nums == null || nums.length == 0) {
            return -1;
        }
        int start = 0;
        int end = nums.length - 1;
        int target = nums[nums.length - 1];

        while (start + 1 < end) {
            int mid = start + (end - start) / 2;
            if (nums[mid] <= target) {
                end = mid;
            }
            else {
                start = mid;
            }
        }
        if (nums[start] <= target) {
            return nums[start];
        } else {
            return nums[end];
        }
    }
}

乱序同字母字符串

给定两个字符串,判断是否颠倒字母而成的字符串,譬如MaryArmy就是同字母而顺序颠倒:

JavaScript

var firstWord = “Mary”; var secondWord = “Army”; isAnagram(firstWord,
secondWord); // true function isAnagram(first, second) { // For case
insensitivity, change both words to lowercase. var a =
first.toLowerCase(); var b = second.toLowerCase(); // Sort the strings,
and join the resulting array to a string. Compare the results a =
a.split(“”).sort().join(“”); b = b.split(“”).sort().join(“”); return a
=== b; }

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var firstWord = "Mary";
var secondWord = "Army";
 
isAnagram(firstWord, secondWord); // true
 
function isAnagram(first, second) {
  // For case insensitivity, change both words to lowercase.
  var a = first.toLowerCase();
  var b = second.toLowerCase();
 
  // Sort the strings, and join the resulting array to a string. Compare the results
  a = a.split("").sort().join("");
  b = b.split("").sort().join("");
 
  return a === b;
}

3# Leetcode 167 Two Sum II – Input array is sorted

/** 
 *  Method one: Two points 一刷
 *    时间复杂度为O(n), 空间复杂度为O(1)。
 */
public int[] twoSum(int[] numbers, int target) {
    int start = 0;
    int end = numbers.length - 1;
    while (start < end) {
        if (numbers[start] + numbers[end] < target) {
            start ++;
        }
        else if(numbers[start] + numbers[end] > target) {
            end --;
        }
        else {
            break;
        }
    }
    return new int[]{start + 1, end + 1};
}

/**
 *     Method 2: Binary Search 一刷
 *     时间复杂度为O(logn), 空间复杂度为O(1)。
 */
public class Solution {
    public int[] twoSum(int[] numbers, int target) {
        int[] result = {0,0};
        int index1 = 0;
        int index2 = 0;

        for(int i = 0; i < numbers.length - 1; i++ ){
            index1 = i + 1;
            if(numbers[i] > target) {
                return result;
            }

            int gap = target - numbers[i];
            int start = i + 1;
            int end = numbers.length - 1;

            while(start + 1 < end){
                int mid = start + (end - start) / 2;
                if(numbers[mid] == gap) {
                    index2 = mid + 1;
                    result[0] = index1;
                    result[1] = index2;
                    return result;
                }
                if (numbers[mid] > gap) {
                    end = mid;
                }
                if (numbers[mid] < gap) {
                    start = mid;
                }
            }
            if (numbers[start] == gap) {
                result[0] = index1;
                result[1] = start + 1;
            }
            if (numbers[end] == gap) {
                result[0] = index1;
                result[1] = end + 1;
            }
        }
       return result;
    }
}

会问字符串

判断某个字符串是否为回文字符串,譬如racecarrace car都是回文字符串:

JavaScript

isPalindrome(“racecar”); // true isPalindrome(“race Car”); // true
function isPalindrome(word) { // Replace all non-letter chars with “”
and change to lowercase var lettersOnly =
word.toLowerCase().replace(/\s/g, “”); // Compare the string with the
reversed version of the string return lettersOnly ===
lettersOnly.split(“”).reverse().join(“”); }

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isPalindrome("racecar"); // true
isPalindrome("race Car"); // true
 
function isPalindrome(word) {
  // Replace all non-letter chars with "" and change to lowercase
  var lettersOnly = word.toLowerCase().replace(/\s/g, "");
 
  // Compare the string with the reversed version of the string
  return lettersOnly === lettersOnly.split("").reverse().join("");
}

24# Leetcode 300. Longest Increasing Subsequence

Given an unsorted array of integers, find the length of longest
increasing subsequence.

For example,
Given [10, 9, 2, 5, 3, 7, 101, 18],
The longest increasing subsequence is [2, 3, 7, 101], therefore the
length is 4. Note that there may be more than one LIS combination, it
is only necessary for you to return the length.

Your algorithm should run in O(n2) complexity.
Follow up: Could you improve it to O(n log n) time complexity?

使用两个栈实现入队与出队

JavaScript

var inputStack = []; // First stack var outputStack = []; // Second
stack // For enqueue, just push the item into the first stack function
enqueue(stackInput, item) { return stackInput.push(item); } function
dequeue(stackInput, stackOutput) { // Reverse the stack such that the
first element of the output stack is the // last element of the input
stack. After that, pop the top of the output to // get the first element
that was ever pushed into the input stack if (stackOutput.length <=
0) { while(stackInput.length > 0) { var elementToOutput =
stackInput.pop(); stackOutput.push(elementToOutput); } } return
stackOutput.pop(); }

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var inputStack = []; // First stack
var outputStack = []; // Second stack
 
// For enqueue, just push the item into the first stack
function enqueue(stackInput, item) {
  return stackInput.push(item);
}
 
function dequeue(stackInput, stackOutput) {
  // Reverse the stack such that the first element of the output stack is the
  // last element of the input stack. After that, pop the top of the output to
  // get the first element that was ever pushed into the input stack
  if (stackOutput.length <= 0) {
    while(stackInput.length > 0) {
      var elementToOutput = stackInput.pop();
      stackOutput.push(elementToOutput);
    }
  }
 
  return stackOutput.pop();
}

22# Leetcode 454. 4Sum II

Given four lists A, B, C, D of integer values, compute how many tuples
(i, j, k, l) there are such that A[i] + B[j] + C[k] + D[l] is
zero.

To make problem a bit easier, all A, B, C, D have same length of N
where 0 ≤ N ≤ 500. All integers are in the range of -228 to 228 – 1
and the result is guaranteed to be at most 231 – 1.

Example:
Input:
A = [ 1, 2]
B = [-2,-1]
C = [-1, 2]
D = [ 0, 2]

Output:
2

Explanation:
The two tuples are:

  1. (0, 0, 0, 1) -> A[0] + B[0] + C[0] + D[1] = 1 + (-2) +
    (-1) + 2 = 0
  2. (1, 1, 0, 0) -> A[1] + B[1] + C[0] + D[0] = 2 + (-1) +
    (-1) + 0 = 0

数组交集

给定两个数组,要求求出两个数组的交集,注意,交集中的元素应该是唯一的。

JavaScript

var firstArray = [2, 2, 4, 1]; var secondArray = [1, 2, 0, 2];
intersection(firstArray, secondArray); // [2, 1] function
intersection(firstArray, secondArray) { // The logic here is to create a
hashmap with the elements of the firstArray as the keys. // After that,
you can use the hashmap’s O(1) look up time to check if the element
exists in the hash // If it does exist, add that element to the new
array. var hashmap = {}; var intersectionArray = [];
firstArray.forEach(function(element) { hashmap[element] = 1; }); //
Since we only want to push unique elements in our case… we can
implement a counter to keep track of what we already added
secondArray.forEach(function(element) { if (hashmap[element] === 1) {
intersectionArray.push(element); hashmap[element]++; } }); return
intersectionArray; // Time complexity O(n), Space complexity O(n) }

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var firstArray = [2, 2, 4, 1];
var secondArray = [1, 2, 0, 2];
 
intersection(firstArray, secondArray); // [2, 1]
 
function intersection(firstArray, secondArray) {
  // The logic here is to create a hashmap with the elements of the firstArray as the keys.
  // After that, you can use the hashmap’s O(1) look up time to check if the element exists in the hash
  // If it does exist, add that element to the new array.
 
  var hashmap = {};
  var intersectionArray = [];
 
  firstArray.forEach(function(element) {
    hashmap[element] = 1;
  });
 
  // Since we only want to push unique elements in our case… we can implement a counter to keep track of what we already added
  secondArray.forEach(function(element) {
    if (hashmap[element] === 1) {
      intersectionArray.push(element);
      hashmap[element]++;
    }
  });
 
  return intersectionArray;
 
  // Time complexity O(n), Space complexity O(n)
}

6# Leetcode 374. Guess Number Higher or Lower

We are playing the Guess Game. The game is as follows:

I pick a number from 1 to n. You have to guess which number I picked.

Every time you guess wrong, I’ll tell you whether the number is higher
or lower.

You call a pre-defined API guess(int num) which returns 3 possible
results (-1, 1, or 0):

-1 : My number is lower
1 : My number is higher
0 : Congrats! You got it!

Example:
n = 10, I pick 6.
Return 6.

/* The guess API is defined in the parent class GuessGame.
   @param num, your guess
   @return -1 if my number is lower, 1 if my number is higher, otherwise return 0
      int guess(int num); */

public class Solution extends GuessGame {
    public int guessNumber(int n) {
        int start = 1, end = n;
        while(start + 1 < end) {
            int mid = start + (end - start) / 2;
            if(guess(mid) == 0) {
                return mid;
            } else if(guess(mid) == 1) {
                start = mid;
            } else {
                end = mid;
            }
        }
        if(guess(start) == 1) {
            return end;
        }
        return start;
    }
}

解释下 null 与 undefined 的区别

JavaScript 中,null 是一个可以被分配的值,设置为 null
的变量意味着其无值。而 undefined
则代表着某个变量虽然声明了但是尚未进行过任何赋值。

33# Leetcode 363. Max Sum of Rectangle No Larger Than K