1. Two Sum

问题：
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].

解答：
这个无非就是时间换空间，空间换时间的计算：
1.时间换空间：

//不能sort, 可能安原位来
public int[] twoSum(int[] nums, int target) {
        int[] result = new int[2];
        if (nums == null || nums.length < 2) return result;
        
        for (int i = 0; i < nums.length - 1; i++) {
            for (int j = i + 1; j < nums.length; j++) {
                if (nums[i] + nums[j] == target) {
                    result[0] = i;
                    result[1] = j;
                    return result;
                }
            }
        }
        
        return result;
    }

2.空间换时间

//快速查找可以用hashmap来做
    public int[] twoSum(int[] nums, int target) {
        int[] result = new int[2];
        if (nums == null || nums.length < 2) return result;
        
        Map<Integer, Integer> map = new HashMap<>();
        for (int i = 0; i < nums.length; i++) {
            if (map.containsKey(target - nums[i])) {
                result[1] = i;
                result[0] = map.get(target - nums[i]);
            }
            map.put(nums[i], i);
        }
        
        return result;
    }