題目：
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].

翻譯：
給定一個(gè)整形數(shù)組和一個(gè)整數(shù)target，返回2個(gè)元素的下標(biāo)，它們滿足相加的和為target。
你可以假定：每次輸入都會(huì)恰好有一個(gè)滿足條件的返回結(jié)果。

（參考別人的解法）
方法一：

    public static int[] twoSum(int[] nums, int target) {

        if(nums == null || nums.length == 0) return null;
        HashMap<Integer, Integer> hashMap = new HashMap<>();
        for(int i = 0; i < nums.length; i++) {
            hashMap.put(nums[i], i);
        }
        int[] result = null;
        for(int j = 0; j < nums.length; j++) {
            int firstNum = nums[j];
            int secondNum = target - firstNum;
            if(hashMap.containsKey(secondNum) && hashMap.get(secondNum) != j) {
                result = new int[2];
                result[0] = j;
                result[1] = hashMap.get(secondNum);
                break;
            }
        }
        return result;      
    }

方法二：

    public static int[] twoSum2(int[] nums, int target) {

        if(nums == null || nums.length == 0) return null;
        int[] result = null;
        for(int i = 0; i < nums.length; i++) {
            for(int j = i+1; j<nums.length; j++) {
                if(nums[i] + nums[j] == target) {
                    result = new int[2];
                    result[0] = i;
                    result[1] = j;
                    break;
                }
            }
        }
        return result;
    }