Given an array of integers, find two numbers such that they add up to a specific target number.

The function twoSum should return indices of the two numbers such that they add up to the target, where index1 must be less than index2. Please note that your returned answers (both index1 and index2) are not zero-based.

You may assume that each input would have exactly one solution.

Input: numbers={2, 7, 11, 15}, target=9
Output: index1=1, index2=2

中文：给定一个int数组，找出两个数，其和是一个特定的目标数。

twoSum函数应该返回这两个数的序号，其中index1必须小于index2。请注意你返回的答案(index1和index2)都不是以0为基准的。

你可以假设每个输入只有一种方案。

输入：numbers[2,7,11,15], target=9

输出：index1=1, index2=2

按照常规的思路，就是数组，首尾同时扫描，求和与target比较，和相等则结束。时间复杂度O(n*n)。这种方法比较慢，提交了之后，服务器那端用1万多长的数组进行测试，结果就是Time Limit Exceeded，超时了。想想也是，若真用数组这样的遍历，就不用出这道题了。

于是改换一种思路，就是用哈希表。把每个数及其index存入到map中，一边存入一边判断，满足条件即返回并终止判断。时间复杂度O(n)。提交即通过，运行时间440 ms。估计还可以再优化。

	//Time Limit Exceeded
	public int[] twoSum(int[] numbers, int target) {
		int[] ret = new int[2];
		for (int j = numbers.length-1; j >= 0; j--) {
			for (int i = 0; i < numbers.length && i<j; i++) {
				if (numbers[i] + numbers[j] == target) {
					ret[0] = i+1;
					ret[1] = j+1;
					break;
				}
			}
		}
		return ret;
	}
	
	public int[] twoSum(int[] numbers, int target){
		int[] ret = new int[2];
		HashMap<Integer,Integer> map = new HashMap<Integer,Integer>();
		for(int i=0;i<numbers.length;i++){
			Integer n = map.get(target - numbers[i]);
			if(n == null)
				map.put(numbers[i], i);
			if(n != null && n < i){
				ret[0] = n+1;
				ret[1] = i+1;
				break;
			}
		}
		return ret;
	}