LeetCode 4Sum C++

#include <iostream> #include <vector> #include<algorithm> #include <queue> using namespace std; /************************************************************************/ /* Problem: Given an array S of n integers, are there elements a, b, c, and d in S such that a + b + c + d = target? Find all unique quadruplets in the array which gives the sum of target. Note: The solution set must not contain duplicate quadruplets. For example, given array S = [1, 0, -1, 0, -2, 2], and target = 0. A solution set is: [ [-1, 0, 0, 1], [-2, -1, 1, 2], [-2, 0, 0, 2] ] Author : crazys_popcorn@126.com DateTime: 2017年8月5日 12:06:17 */ /************************************************************************/ class Solution { public: vector<vector<int>> fourSum(vector<int>& nums, int target) { vector< vector <int > >_result; int _size = nums.size(); if (nums.size() < 4) return _result; sort(nums.begin(), nums.end()); for (int i = 0; i < _size - 3; i++) { if (i>0 && nums[i]== nums[i-1]) continue; if (nums[i]+nums[i+1]+nums[i+2] +nums[i+3]> target) break; if (nums[i] + nums[_size - 1] + nums[_size - 2] + nums[_size - 3] < target) continue; for (int j = i + 1;j <_size - 2; j++) { //为什么要用j > i+1 因为这是 要往后推迟一个。不要让他和i相比 也就是前一位数字 if ( j > i +1 &&nums[j] == nums[j - 1]) continue; if (nums[i] + nums[j] + nums[j+1] + nums[j+2]> target) break; if (nums[i] + nums[_size - 1] + nums[_size - 2] + nums[j] < target) continue; int k = j + 1, m = _size - 1; while (k < m) { int sum = nums[i] + nums[j] + nums[k] + nums[m]; if (sum <target)k++; else if (sum > target) m--; else { _result.push_back(vector<int>{ nums[i], nums[j], nums[k], nums[m] }); do{ k++;} while (nums[k] == nums[k - 1] && k < m); do{ m--;} while (nums[m] == nums[m + 1] && k < m); } } } } return _result; } }; void main() { Solution s1; // 0 1 2 3 4 5 6 7 vector<int> _temp = { -5, -5, -3, -1, 0, 2, 4, 5 }; vector<vector<int>> num = s1.fourSum(_temp, -11); std::cout << "" << endl; system("pause"); }