Two Sum

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.

Example:
Given nums = [2, 7, 11, 15], target = 9,
Because nums[0] + nums[1] = 2 + 7 = 9,
return [0, 1].

解法一（暴力破解法:77ms）：
两重循环，第一重循环遍历数组，第二重循环从上一重下标的下一个下标开始遍历数组，如果某次两个下标指示的数相加等于目标值，此时返回两重循环的下表即可。

public class Solution {    public int[] twoSum(int[] nums, int target) {        int len = nums.length;        for(int i = 1;i < len;i++){            for(int j = 0;j < i;j++){                int temp = nums[i]+nums[j];                if(temp == target) return new int[]{j,i};            }        }        //这里是系统异常，所以不需要throws        throw new IllegalArgumentException("No two sum solution");    }}

解法二（HashMap:6ms）：
遍历过程中不断把之前的数据添加到HashMap中，每一次检查HashMap有没有正好等于Target减去当前值的值，有则返回下标。

public class Solution {    public int[] twoSum(int[] nums, int target){       Map<Integer,Integer> a = new HashMap<Integer,Integer>();       for(int i = 0;i < nums.length;i++){           if(a.containsKey(target-nums[i]))                return new int[]{a.get(target-nums[i]),i};            else a.put(nums[i],i);       }       throw new IllegalArgumentException("No two sum solution");    }}