两数之和 two-sum

题目描述

给出一个整数数组，请在数组中找出两个加起来等于目标值的数，
你给出的函数twoSum 需要返回这两个数字的下标（index1，index2），需要满足 index1 小于index2.。注意：下标是从1开始的
假设给出的数组中只存在唯一解
例如：
给出的数组为 {2, 7, 11, 15},目标值为9
输出 ndex1=1, index2=2

Given an array of integers, find two numbers such that they add up to a specific target number.
The function twoSum should return indices of the two numbers such that they add up to the target, where index1 must be less than index2. Please note that your returned answers (both index1 and index2) are not zero-based.

You may assume that each input would have exactly one solution.

Input: numbers={2, 7, 11, 15}, target=9
Output: index1=1, index2=2

示例

输入

[3,2,4],6

输出

[2,3]

解题思路

• hash表，用unordered_map记录“元素值->数组下标”的映射

#include <unordered_map>
class Solution {
    
    
public:
    vector<int> twoSum(vector<int>& numbers, int target) {
    
    
        vector<int> res;
        unordered_map<int, int> mp;
        int size = numbers.size();
        //初始化hash表
        for(int i = 0; i < size; i ++) {
    
    
            mp[numbers[i]] = i;
        }
        for(int i = 0; i < size; i ++) {
    
    
            int temp = target - numbers[i];
            //如果在mp中找到互补的元素，则添加进结果
            //同时要确保互补的数不是它本身
            if(mp.find(temp) != mp.end() && mp[temp] > i) {
    
    
                res.push_back(i + 1);  //由于下标是从1开始的，所以要加一
                res.push_back(mp[temp] + 1);
                break;
            } else {
    
    
                //如果在mp中没找到互补的元素，则直接跳过
                continue;
            }
        }
        return res;
    }
};