1,Two Sum

1，题目：
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].
UPDATE (2016/2/13):
The return format had been changed to zero-based indices. Please read the above updated description carefully.

2，思路：
使用HashMap用来存储（数组中大值作为key，对应下标作为value）每次遍历到一个元素，将target减去该数组元素值，看其是否在HashMap中即可。
时间复杂度O(n)
空间复杂度O(n)

3，代码：

public static int[] twoSum(int[] nums, int target) {        if (nums == null || nums.length <= 2) {            return null;        }        //数组中的值作为key，对应的下标作为value        Map<Integer, Integer> map = new HashMap<Integer, Integer>();        for (int i = 0, len = nums.length; i < len; i++) {            if(map.get(target-nums[i]) != null){                int[] result = {map.get(target-nums[i]),i};                return result;            }            map.put(nums[i], i);        }        return new int[2];    }