LeetCode Oj 15. 3Sum_3sum [difficulty: medium]-优快云博客

本文链接：https://blog.youkuaiyun.com/u014737310/article/details/51345368

本文探讨了如何寻找数组中三个元素相加等于零的所有唯一组合，并提供了一个C++实现方案。通过排序和双指针技巧，有效地解决了重复元素带来的问题。

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15. 3Sum

Total Accepted: 116381 Total Submissions: 617430 Difficulty: Medium

Given an array S of n integers, are there elements a, b, c in S such that a + b + c = 0? Find all unique triplets in the array which gives the sum of zero.

Note:

Elements in a triplet (a,b,c) must be in non-descending order. (ie, a ≤ b ≤ c)
The solution set must not contain duplicate triplets.

    For example, given array S = {-1 0 1 2 -1 -4},

    A solution set is:
    (-1, 0, 1)
    (-1, -1, 2)

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class Solution {
public:
    vector<vector<int> > threeSum(vector<int>& nums) {


        vector<vector<int> >ans;
        sort(nums.begin(),nums.end());
        for(int i=0;i<nums.size();i++)
        {
            int iii=i;
            while(i>0&&i<nums.size()&&nums[i]==nums[i-1]) i++;

            if(i>=nums.size()) break;
            int target=-nums[i];
            int s,e;
            s=i+1;
            e=nums.size()-1;
            int ss=s,ee=e,ii=i;

            while(s<e)
            {

               if(nums[s]+nums[e]==target)
               {
                   vector<int>temp;
                   temp.push_back(nums[i]);
                   temp.push_back(nums[s]);
                   temp.push_back(nums[e]);
                   ans.push_back(temp);
                   s++;
                   while(s<e&&nums[s]==nums[s-1]) s++;
               }

               while(s<e && (nums[s]+nums[e]<target)) s++;
               while(s<e && (nums[s]+nums[e]>target)) e--;
            }
        }
        return ans;
    }
};