Leetcode -- 3sum

Problem description:

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)

Analysis: The question requires that all possible sets be found. Therefore, the thought is to sort the array first, and then use two repeated loops to select two numbers A and B in sequence, then, find whether there is C in the remaining number. The specific implementation uses the upper_bound function to find the first number larger than the current number and remove the repeating loop, then, use the find function to check whether C exists. If C exists, the three numbers are recorded. The Code is as follows:

class Solution {public:    vector<vector<int> > threeSum(vector<int> &num) {        vector<vector<int> > results;        if(num.size()<3) return results;        vector<int> subset;        sort(num.begin(),num.end());        vector<int>::iterator p=num.begin(),q,flag;        while(p<(num.end()-2))        {            q=p+1;            while(q<num.end()-1)            {                int tag=0-*p-*q;                if(find(q+1,num.end(),tag)!=num.end())                {                flag=find(q+1,num.end(),tag);                subset.push_back(*p);                subset.push_back(*q);                subset.push_back(*flag);                results.push_back(subset);                }                subset.clear();                q=upper_bound(q,num.end()-1,*q);            }            p=upper_bound(p,num.end()-2,*p);        }        return results;    }};