LeetCode 1. Two Sum

发表于 | 分类于 |

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:

1
2
3
4
Given nums = [2, 7, 11, 15], target = 9,

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

思路:
假设每个输入都只有一个解决方案，先新建一个Map集合，先循环一遍nums数组，以target - nums[i]为key，以索引i为value，
放入Map中；然后再循环一遍nums数组，以nums[i]为key，取出v，如果v存在且不等与i，则代表找到了匹配值，返回唯一解int[]{i, v}。

Solution:

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
public class Solution {
    public int[] twoSum(int[] nums, int target) {

        Map<Integer, Integer> m = new HashMap<Integer, Integer>();

        for(int i = 0; i < nums.length; i++){
            m.put(target - nums[i], i);
        }

        for(int i = 0; i < nums.length; i++){

            Integer v = m.get(nums[i]);

            if(v != null && v != i){
                return new int[]{i, v};
            }
        }

        throw new RuntimeException();
    }
}