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].

Method 1. brutal, O(n^2).
class Solution {
    public int[] twoSum(int[] nums, int target) {
        int[] res = new int[2];
        if(nums == null || nums.length == 0) return res;
        for(int i = 0; i < nums.length; i++) {
            for(int j = i + 1; j < nums.length; j++) {
                if(nums[i] + nums[j] == target) {
                    res[0] = i;
                    res[1] = j;
                    return res;
                }
            }
        }
        return null;
    }
}

Method 2. use hashmap, O(n).
class Solution {
    public int[] twoSum(int[] nums, int target) {
        int[] res = new int[2];
        if(nums == null || nums.length < 2) return res;

        HashMap<Integer, Integer> map = new HashMap<>();
        for(int i = 0; i < nums.length; i++) {
            if(map.containsKey(target - nums[i])) {
                res[0] = map.get(target - nums[i]);
                res[1] = i;
                return res;
            }
            map.put(nums[i], i); //判断后再放map里，要不[3, 3]这样会出问题
        }
        return null;
    }
}