JavaScript Two Sum

TwoSum-LeetCode

LeetCode

Publish Date: 2019-04-25

Read Count:

Given an array of integers, return indices of the two numbers such that they add up to a specific target.

You may assume that each input would have exactly one solution, and you may not use the same element twice.

给定一个整数数组，返回两个数字的索引，使它们相加到特定目标。您可以假设每个输入只有一个解决方案，并且您可能不会两次使用相同的元素。

Example:

Given nums = [2, 7, 11, 15], target = 9,

Because nums[0] + nums[1] = 2 + 7 = 9,
return [0, 1].

解法一双重for循坏暴力破解

/**
 * @param {number[]} nums
 * @param {number} target
 * @return {number[]}
 */
var twoSum = function(nums, target) {
    let result = [];
    for (let i = 0; i < nums.length; i++) {
        for (let j = i + 1; j < nums.length; j++) {
           if (nums[i] + nums[j] === target) {
            result.push(i);
            result.push(j);
        }
      }
   }
    return result;
};

❤ 解法一双重for循坏暴力破解Runtime:160ms,Memory:34.7MB,时间复杂度是O(n^2),暴力破解时间复杂度高

暴力破解for循坏

function hasSum(arr, k) {
  hashMap = {};
  for (let value of arr) {
    if (hashMap[value]) {
      return true;
    } else {
      hashMap[k - value] = true
    }
    return false
  }
}

使用ES6的数据结构Map

使用ES6的数据结构Map时间复杂度是O(n)

/**
 * @param {number[]} nums
 * @param {number} target
 * @return {number[]}
 */
var twoSum = function (nums, target) {
  let map = new Map();
  for (let i = 0; i < nums.length; i ++) {
    return [map.get(target -nums[i]), i];
  } else {
    map.set[nums[i], i]
  }
  return []
}