LeetCode(Two Sum)

最新推荐文章于 2024-03-02 21:28:00 发布

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最新推荐文章于 2024-03-02 21:28:00 发布

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分类专栏： LeetCode 面试题 c++ 文章标签： leetcode 面试

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本文介绍了一种解决两数之和问题的有效算法。给定一个整数数组及目标值，寻找两个数相加等于该目标值的下标。算法首先对数组进行排序并使用双指针从两端逼近，找到匹配的元素后，在原始数组中确定它们的位置。

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题目要求：

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

注意数组是没有排序的，输出的索引是原始位置例如 3，2，4 target = 6 输出： 2， 3

思路：先将原始数组复制到临时数组然后一前一后指针查找，找到后再在原数组中找到原始位置。

class Solution {
public:
    
    vector<int> twoSum(vector<int> &numbers, int target) {
        vector<int> tmp(numbers);
        sort(tmp.begin(), tmp.end());
        int idx1 = 0, idx2 = numbers.size() - 1;
        while(idx1 < idx2)
        {
            
            if(target < (tmp[idx1] + tmp[idx2]))
                --idx2;
            else if(target > (tmp[idx1] + tmp[idx2]))
                ++idx1;
            else
                break;
        }
        vector<int> ret;
        int a = tmp[idx1], b = tmp[idx2];
        idx1 = -1, idx2 = -1;
        for (size_t i = 0; i < numbers.size(); ++i) {
            if (numbers[i] == a || numbers[i] == b) {
                if(idx1 == -1)
                    idx1 = i + 1;
                else
                    idx2 = i + 1;
            }
        }
        if(idx1 > idx2)
            swap(idx1, idx2);
        ret.push_back(idx1);
        ret.push_back(idx2);
        return ret;
    }
};

代码：