leetcode--Two Sum

最新推荐文章于 2025-08-21 10:38:03 发布

转载最新推荐文章于 2025-08-21 10:38:03 发布 · 46 阅读

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原文链接：http://www.cnblogs.com/obama/p/3266680.html

文章标签：

#数据结构与算法 #c/c++

本文介绍了一种使用哈希表快速解决数组中寻找两数之和等于特定目标值的问题的方法，包括特殊情况处理及代码实现。

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1.题目描述

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

1.解法分析

由于题目说得很明白，一定有解且解只有一组，那么这题用hash做自然是最快了，但是对于偶数输入，比如target为8，如果结果是4和4的相加需要特殊考虑。

class Solution {

public:

    vector<int> twoSum(vector<int> &numbers, int target) {

        // Start typing your C/C++ solution below

        // DO NOT write int main() function

        unordered_set<int> myHash;

        vector<int>::iterator iter;

        vector<int> result;

        result.assign(2,0);

        if(target%2==0)

            int count=0;

            for(int i=0;i<numbers.size();++i)

                if(numbers[i]==target/2)

                    if(count==0){result[0]=i+1;count++;}

                    else {result[1]=i+1;return result;}

        for(iter=numbers.begin();iter!=numbers.end();++iter)

            myHash.insert(*iter);

        for(int i=0;i<numbers.size();++i)

            myHash.erase(numbers[i]);

            if(myHash.count(target-numbers[i])==1)

                result[0]=i+1;break;

            myHash.insert(numbers[i]);

        for(int i=result[0];i<numbers.size();++i)

            if((numbers[result[0]-1]+numbers[i])==target)

                result[1]=i+1;break;

        return result;

};

转载于:https://www.cnblogs.com/obama/p/3266680.html