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最新推荐文章于 2022-03-02 23:29:33 发布

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最新推荐文章于 2022-03-02 23:29:33 发布

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本文链接：https://blog.youkuaiyun.com/Matthew_lk/article/details/50491680

本文探讨了解决TwoSum问题的两种方法，包括时间复杂度为O(N^2)的暴力搜索法和更高效的解决方案。重点介绍了如何通过哈希表优化算法性能，避免重复计算并显著提高效率。

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1.Two Sum

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

题目大意：给定一个数组，找出数组中两个数相加等于目标值，返回这两个数的索引值。

方法一：

class Solution {
public:
    vector<int> twoSum(vector<int>& nums, int target) {
        
        int i,j, sum;
        vector<int> results;
        
        for(i=0;i<(nums.size()-1);i++)
        {
            for(j=i+1;j<nums.size();j++){
                sum = nums[i]+nums[j];
                if(sum == target)
                {
                    results.push_back(i+1);
                    results.push_back(j+1);
                }
            }
        }
        return results;
    }
};

该方法时间复杂度O（N^2）,代码提交超时。

方法二：