前言
Given an array of integers nums and an integer target, return indices of the two numbers such that they add up to target.
You may assume that each input would have exactly one solution, and you may not use the same element twice.
You can return the answer in any order.
来源：力扣（LeetCode） 链接：https://leetcode-cn.com/problems/add-two-numbers
题目
Example 1:
代码语言：javascript复制Input: nums = [2,7,11,15], target = 9
Output: [0,1]
Output: Because nums[0] nums[1] == 9, we return [0, 1]Example 2:
代码语言：javascript复制Input: nums = [3,2,4], target = 6
Output: [1,2]Example 3:
代码语言：javascript复制Input: nums = [3,3], target = 6
Output: [0,1]Constraints:
代码语言：javascript复制2 <= nums.length <= 103
-109 <= nums[i] <= 109
-109 <= target <= 109
Only one valid answer exists.解题
PHP
代码语言：javascript复制class Solution {
/**
* @param Integer[] $nums
* @param Integer $target
* @return Integer[]
*/
function twoSum($nums, $target) {
//定义数组存放下标值
$indices = [];
foreach ($nums as $key => $val) {
//得到差异值
$diff = $target - $val;
//判断差异值是否在数组$indices中
//如果不在数组$indices,就将key存入$indices数组中,继续寻找
//若在数组则直接返回
if (!isset($indices[$diff])) {
$indices[$val] = $key;
continue;
}
return [$indices[$diff], $key];
}
}
}JavaScript
代码语言：javascript复制/**
* @param {number[]} nums
* @param {number} target
* @return {number[]}
*/
var twoSum = function(nums, target) {
//创建一个map对象
const map = new Map();
//同PHP思路一致
for(var i= 0; i<nums.length; i ) {
let diff = target - nums[i];
if (!map.has(diff)) {
map.set(nums[i],i)
continue;
}
return [map.get(diff), i];
}
};
