【LeetCode】001.Two Sum

Description：

Given an array of integers, return indices of the two numbers such that they add up to a specific target.
给定一个整数数组，返回两个数的索引，使它们相加为一个特定的目标。

You may assume that each input would have exactly one solution, and you may not use the same element twice.
可以假设每个输入都有一个解决方案，但是同一个元素不能使用两次。

Example:

Given nums = [2, 7, 11, 15], target = 9,

Because nums[0] + nums[1] = 2 + 7 = 9,
return [0, 1].

我的解决方法：

public class Solution {
    public int[] twoSum(int[] nums, int target) {
        int[] res = new int[2];
        for (int i = 0; i < nums.length; i++) {
            for (int j = i+1; j < nums.length; j++) {
                if (target-nums[i]==nums[j]) {
                    res[0] = i;
                    res[1] = j;
                    break;
                }
            }
        }
        return res;
    }
}

DS方法：

public class Solution {
    public int[] twoSum(int[] numbers, int target) {
        HashMap<Integer,Integer> hash = new HashMap<Integer,Integer>();
        for(int i = 0; i < numbers.length; i++){
            Integer diff = (Integer)(target - numbers[i]);
            if(hash.containsKey(diff)){
                int toReturn[] = {hash.get(diff)+1, i+1};
                return toReturn;
            }
            hash.put(numbers[i], i);
        }
        return null;
    }
}