[Leetcode]之一《two sum》解题报告-优快云博客

本文链接：https://blog.youkuaiyun.com/Kid_U_ForFun/article/details/20286065

本文探讨了在整数数组中寻找两个数使它们的和等于特定目标值的问题，并提供了一种有效的解决方案。该方法利用了类似哈希的技术来减少搜索时间，通过一个示例代码展示了具体的实现过程。

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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

题目要求在数组中找到两个数，满足其和为target，输出其下标。

有三种方法：一个一个的找、排序从中部向两头找（有点像归并排序的翻版）、类似哈希的办法（使用了map）。一看网上说第一种会超时，然后第二种办法太麻烦了，就选了第三种办法，竟然不会写，抄了个代码，中间漏洞百出，竟然是全角符号和半角符号的问题。

代码如下（抄的，来自http://blog.youkuaiyun.com/ithomer/article/details/8783952）：

class Solution {
    public:
     vector<int> twoSum(vector<int> &numbers, int target){
         
          int len = numbers.size();
          assert(len >= 2);
          
          vector<int> ret(2, 0);
          
          map<int, int> mapping;
          vector<long long> mul(len, 0);
          
          for(int i = 0;i < len; i++){
              mul[i] = (target - numbers[i]) * numbers[i];
              
              if(mapping[mul[i]] > 0){
                  if(numbers[i] + numbers[mapping[mul[i]] - 1] == target){
                      ret[0] = mapping[mul[i]];
                      ret[1] = i + 1;
                      break;
                  }
              } else{
                  mapping[mul[i]] = i + 1;
              }
          }
          return ret;
     }
};