LeetCode：Two Sum

最新推荐文章于 2025-08-04 20:35:07 发布

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最新推荐文章于 2025-08-04 20:35:07 发布

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分类专栏： LeetCode 文章标签： leetcode c++ c语言

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本文介绍了一种解决两数之和问题的算法实现。给定整数数组及目标值，寻找两个数相加等于目标值的下标。通过排序与双指针技术，文章详细展示了如何高效找到符合条件的元素及其原始下标。

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问题描述：

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：

code：

class Solution {
public:
    vector<int> twoSum(vector<int>& nums, int target) {
    int index = 0;
    int index1 = 0;
    int index2 = 0;
    vector<int> nums1(nums);
    sort(nums1.begin(),nums1.end());   //升序排序
    vector<int>::size_type length = nums1.size();
    cout<<length<<endl;
    for(vector<int>::size_type ix = 0,is = --length;is != ix;)
    {
index = nums1[ix] + nums1[is];
if(index == target)
{
            index1 = ++ix;
            index2 = ++is;
<span style="white-space:pre">	</span>}
<span style="white-space:pre">	</span>else  if(index < target)
              { 
 <span style="white-space:pre">	</span>           ++ix;
 <span style="white-space:pre">	</span>      }
              else  
 <span style="white-space:pre">	</span>      { 
        <span style="white-space:pre">	</span>   --is; 
 <span style="white-space:pre">	</span>      } 
    }
   cout<<"index1 = "<<index1<<","<<"index2 = "<<index2<<endl;
   vector<int>::size_type index3 = 0;
   vector<int>::size_type index4 = 0;
   for(vector<int>::size_type ix = 0; ix != length; ++ix )
   {
       if(nums[ix] == index1)
          index3 = ++ix;
       if(nums[ix] == index2)
          index4 = ++ix;
   }
   cout<<"index3 = "<<index3<<","<<"index4 = "<<index4<<endl;
    }
};