LeetCode_三数之和

#include<iostream>
#include <vector>
#include <set>
#include <functional>
#include <stdlib.h>
#include <stdio.h>
#include <string>
#include <sstream>
#include <list>
#include <map>
#include <stack>
#include <algorithm>
//#include <iterator>
using namespace std;

#include <iostream>
#include <string>
using namespace std;


vector<vector<int>> threeSum(vector<int>& nums) {
    int length = nums.size();
    if (length < 3)
        return vector<vector<int>>();
    int pa = 0, pb = 1, pc = length - 1;
    sort(nums.begin(), nums.end(), [](const int &a, const int &b){return a < b; });

    vector<vector<int>> result;
    for (int pa = 0; pa < length - 2; pa++)
    {
        pb = pa + 1;
        pc = length - 1;
        int value = -nums[pa];
        while (pb < pc)
        {
            if (nums[pb] + nums[pc] > value)
                pc--;
            else if (nums[pb] + nums[pc] < value)
                pb++;
            else
            {
                result.push_back({ nums[pa], nums[pb], nums[pc] });
                while (nums[pb] == nums[pb+1])
                    pb++;
                while (nums[pc] == nums[pc - 1])
                    pc--;
                pb++;
                pc--;
            }
        }
    }
    sort(result.begin(), result.end());
    auto end = unique(result.begin(), result.end());
    result.erase(end, result.end());
    return result;
}
//void unique(vector<vector<int>> &group)
//{
//
//}
int main()
{
    vector<int> vi = { -4, -2, -2, -2, 0, 1, 2, 2, 2, 3, 3, 4, 4, 6, 6 };
    vector<vector<int>> result = threeSum(vi);

    for (auto v : result)
    {
        for (auto i : v)
            cout << i << " ";
        cout << endl;
    }


    return 0;
}

正常需要o(n^3)复杂度，可采用先固定第一位，然后后两位用游标的方式，复杂度(n^2)