leecode 解题总结：18 4Sum

#include <iostream>#include <stdio.h>#include <vector>#include <algorithm>#include <set>using namespace std;/*问题: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.分析:用3Sum来做。罗列第一个数和第二个数，时间复杂度为O(n^3)A solution set is:[  [-1,  0, 0, 1],  [-2, -1, 1, 2],  [-2,  0, 0, 2]输入:6(数组元素个数) 0(目标和)1 0 -1 0 -2 29 120 4 -5 2 -2 4 2 -1 4输出:-1 0 0 1, -2 -1 1 2, -2 0 0 20 4 4 4, 2 2 4 4关键:1 4Sum用3Sum来做。罗列第一个数和第二个数，时间复杂度为O(n^3)for(int i = 0 ; i < size - 2 ; i++){int target_3 = target - nums.at(i);for(int j = i + 1 ; j < size - 1; j++){int target_2 = target_3 - nums.at(j);int low = j + 1;int high = size - 1;while(low < high){sum = nums.at(low) + nums.at(high);if(sum < target_2){low++;}else if(sum > target_2){high--;}else{Result result(nums.at(i) , nums.at(j) , nums.at(low) , nums.at(high));setResult.insert(result);low++;high--;}}}}*/typedef struct Result{Result(){}Result(int min ,int mid1 , int mid2 , int max):_min(min),_mid1(mid1),_mid2(mid2),_max(max){}bool operator < (const Result& result) const{if(_min != result._min){return _min < result._min;}else if(_mid1 != result._mid1){return _mid1 < result._mid1;}else if(_mid2 != result._mid2){return _mid2 < result._mid2;}else{return _max < result._max;}}int _min;int _mid1;//次最小int _mid2;//次最大int _max;}Result;class Solution {public:vector<vector<int>> fourSum(vector<int>& nums, int target){vector<vector<int>> results;if(nums.empty() || nums.size() < 4){return results;}int size = nums.size();sort(nums.begin() , nums.end());int sum;set<Result , less<Result> > setResult;//这里的关键是罗列的时候:我已经实现罗列的a<=b,后续确定的另外两个数c,d有:a <= b <= c <= d//这样罗列比如:a=A[1], b=A[3],那么是否c可以为A[2]?不可以，已经假定了顺序。那么也不会存在判重的问题for(int i = 0 ; i < size - 2 ; i++){int target_3 = target - nums.at(i);for(int j = i + 1 ; j < size - 1; j++){int target_2 = target_3 - nums.at(j);int low = j + 1;int high = size - 1;while(low < high){sum = nums.at(low) + nums.at(high);if(sum < target_2){low++;}else if(sum > target_2){high--;}else{Result result(nums.at(i) , nums.at(j) , nums.at(low) , nums.at(high));setResult.insert(result);low++;high--;}}}}//返回结果for(set<Result , less<Result> >::iterator it = setResult.begin() ; it != setResult.end() ; it++){vector<int> vecResult;//vector<int>{nums[i],nums[j],nums[left],nums[right]}vecResult.push_back(it->_min);vecResult.push_back(it->_mid1);vecResult.push_back(it->_mid2);vecResult.push_back(it->_max);results.push_back(vecResult);}return results;}};void print(vector< vector<int> >& results){if(results.empty()){cout << "no result" << endl;return;}int size = results.size();for(int i = 0 ; i < size ; i++){vector<int> result = results.at(i);int len = result.size();for(int j = 0; j < len ; j++){cout << result.at(j) << " ";}cout << ",";}cout << endl;}void process(){int num;vector<int> nums;vector< vector<int> > results;int value;int target;while(cin >> num >> target){nums.clear();for(int i = 0 ; i < num ; i++){cin >> value;nums.push_back(value);}Solution solution;results = solution.fourSum(nums , target);print(results);}}int main(int argc , char* argv[]){process();getchar();return 0;}