Leetcode sum of three numbers c++

Given an array nums containing n integers, determine whether there are three elements a, b, c in nums such that a + b + c = 0? Find all triples that meet the conditions and are not repeated.

Note: The answer cannot contain repeated triples.

For example, given the array nums = [-1, 0, 1, 2, -1, -4],

The set of triples that meet the requirements are:
[
[-1, 0, 1],
[-1, -1, 2]
]

---

Solution 1: Violence

class Solution {
public:
    vector<vector<int>> threeSum(vector<int>& nums) {
        if(nums.size()<3) return{};
        vector<vector<int>> res;//结果
        set<vector<int>> ret;//用来去重
        for(int i=0;i<nums.size()-2;i++)
        {
            for(int j=i+1;j<nums.size()-1;j++)
            {
                for(int k=j+1;k<nums.size();k++)
                {
                    if(nums[i]+nums[k]+nums[j]==0)
                    {
                        vector<int> temp;
                        int a=(nums[i]<nums[j]?nums[i]:nums[j])<nums[k]?(nums[i]<nums[j]?nums[i]:nums[j]):nums[k];//放最小的元素
                        int b=(nums[i]>nums[j]?nums[i]:nums[j])>nums[k]?(nums[i]>nums[j]?nums[i]:nums[j]):nums[k];//放最大的元素
                        int c=0-a-b;
                        temp.push_back(a);
                        temp.push_back(c);
                        temp.push_back(b);
                        ret.insert(temp);
                    }
                }
            }
        }
        for(auto it:ret)
        {
            res.push_back(it);
        }
        return res;
    }
};

The time complexity is O(n ³ ), and the 282th use case will time out

Solution 2: Optimization of violence law

Find directly whether the third number exists

class Solution {
public:
    static bool cmp(const int&a,const int &b)
    {
        return a<b;
    }
    vector<vector<int>> threeSum(vector<int>& nums) {
        sort(nums.begin(),nums.end(),cmp);
        vector<int>::iterator it;
        vector<vector<int>> res;
        set<vector<int>> a;
        for(int i=0;i<nums.size();i++)
        {
            if(nums[i]>0) break;
            for(int j=i+1;j<nums.size();j++)
            {
                it=find(nums.begin()+j+1,nums.end(),0-nums[i]-nums[j]);
                if(it!=nums.end())
                {
                    vector<int> temp;
                    temp.push_back(nums[i]);
                    temp.push_back(nums[j]);
                    temp.push_back(*it);
                    a.insert(temp);
                }
            }
        }
        for(auto k:a)
        {
            res.push_back(k);
        }
        return res;
    }
};

The efficiency has been improved to a certain extent, but it will still time out in 311 use cases

Solution 3: 3-pointer method

class Solution {
public:
    vector<vector<int>> threeSum(vector<int>& nums) {
        sort(nums.begin(),nums.end());
        if(nums.size()<3||nums.front()>0||nums.back()<0) return {};
        vector<vector<int>> res;
        for(int i=0;i<nums.size();i++)
        {
            int fix=nums[i];
            if(fix>0) break;
            if(i>0&&fix==nums[i-1])
                continue;
            int l=i+1;
            int r=nums.size()-1;
            while(l<r)
            {
                if(nums[l]+nums[r]==-fix)
                {
                    if(l==i+1 || r==nums.size()-1)
                    {
                        res.push_back(vector<int>{nums[i],nums[l],nums[r]});
                        l++;
                        r--;
                    }
                    else if(nums[l]==nums[l-1])
                        l++;
                    else if(nums[r]==nums[r+1])
                        r--;
                    else
                    {
                        res.push_back(vector<int>{nums[i],nums[l],nums[r]});
                        l++;
                        r--;
                    }
                }
                else if(nums[l]+nums[r]<-fix)
                    l++;
                else r--;
            }
        }
        return res;
    }
};