link: https://leetcode.com/problems/two-sum/description/
> 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.
Example:
```
Given nums = [2, 7, 11, 15], target = 9,
Because nums[0] + nums[1] = 2 + 7 = 9,
return [0, 1].
```
# JavaScript
```js
/**
* @param {number[]} nums
* @param {number} target
* @return {number[]}
*/
```
>[danger] runtime: 7056 ms
```js
var twoSum = function(nums, target) {
for (let i in nums) {
for (let j in nums) {
if (i !== j && nums[i] + nums[j] === target) {
return [parseInt(i), parseInt(j)]
}
}
}
return [-1, -1]
};
```
>[danger] runtime: 312 ms
```js
var twoSum = function(nums, target) {
for (let i = 0; i < nums.length; i++) {
for (let j = 0; j < nums.length; j++) {
if (i !== j && nums[i] + nums[j] === target) {
return [parseInt(i), parseInt(j)]
}
}
}
return [-1, -1]
};
```
>[success] runtime: 82 ms
```js
var twoSum = function(nums, target) {
const dict = {};
for (var i = nums.length; i >= 0 ; i--) {
if (dict[nums[i]] !== undefined) {
return [dict[nums[i]], i];
}
dict[target - nums[i]] = i;
}
};
```
>[success] runtime: 86 ms
```js
var twoSum = function(nums, target) {
var index = {}
for(var i = 0,len = nums.length; i < len; i++) {
var matchNum = target - nums[i];
if(typeof index[matchNum] === 'number') return [index[matchNum], i]
index[nums[i]] = i;
}
};
```
> 總結: 字典查詢速度特別快，盡量少用`for-in`循環，普通`for`循環速度一般，多級循環會降低執行速度。
