LeetCode之4Sum（经典）

题目：

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:
• Elements in a quadruplet (a,b,c,d) must be in non-descending order. (ie, a ≤ b ≤ c ≤ d)
• 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)

分析：给定一个包含n个数的数组，问是否在数组中存在四个数abcd，使得a+b+c+d = target。在数组中找出所有的符合条件的四个数的集合。要求每个集合不重复，并且abcd从小到大排列。方法三种：

1.先排序，然后左右夹逼,时间复杂度 O(n^3)，空间复杂度 O(1)
2.先排序然后缓存两个数的和，最后，遍历。
3.先排序，然后缓存两个数的和，然后对这些和进行2sum。要求坐标不等。

代码：

#include "stdafx.h"#include <ctime>

#include <iostream>#include <vector>#include <algorithm>#include <unordered_map>using namespace stdext;using namespace std;

class Solution{public:vector< vector<int> > FourSum1(vector<int>& num, int target){vector<vector<int>> result;vector<int> temp(4);sort(num.begin(),num.end());auto last = num.end();for (auto a = num.begin();a != last - 3;++a){for(auto b = next(a);b != prev(last,2);++b){auto c = next(b);auto d = prev(last);while(c < d){if (*a + *b + *c + *d < target){c++;}else if (*a + *b + *c + *d >target){ d--;}else{temp[0]=*a;  temp[1]=*b;temp[2]=*c;  temp[3]=*d;result.push_back(temp);++c;--d;}}}}sort(result.begin(), result.end());//unique移除相邻重复的元素，然后返回的是新的结尾指针，erase擦除多余的内存部分。result.erase(unique(result.begin(),result.end()), result.end());return result;}vector< vector<int> > FourSum2(vector<int>& num, int target){vector<vector<int>> result;if (num.size() < 4) return result;vector<int> temp(4);sort(num.begin(),num.end());unordered_map<int, vector<pair<int, int>> > twoSum; //重复的键值，将其存放在一个vector中auto last = num.end();for (size_t a=0;a<num.size();++a){for (size_t b = a+1;b<num.size();++b){twoSum[num[a] + num[b]].push_back(pair<int,int>(num[a],num[b]));}}int sub = 0;for(size_t c=0;c<num.size();++c){for (size_t d=c+1;d<num.size();++d){sub = target - num[c] - num[d];if(twoSum.find(sub) == twoSum.end())continue;auto& subRst = twoSum[sub];//可能有多个结果。for (size_t i=0;i<subRst.size();++i){if(num[c] <= subRst[i].first || num[d] <= subRst[i].second) //去除重复的情况，pair中的元素从小到大排序，所以secondcontinue;temp[0] = subRst[i].first;temp[1] = subRst[i].second;temp[2] = num[c];temp[3] = num[d];sort(temp.begin(),temp.end());result.push_back(temp);}}}sort(result.begin(), result.end());result.erase(unique(result.begin(),result.end()), result.end());return result;}vector< vector<int> > FourSum3(vector<int>& num, int target){vector<vector<int>> result;if (num.size() < 4) return result;vector<int> temp(4);sort(num.begin(),num.end());unordered_multimap< int,pair<int, int> > twoSum;for(size_t i = 0;i<num.size();++i){for (size_t j=i+1;j<num.size();++j){twoSum.insert(make_pair(num[i]+ num[j], make_pair(num[i],num[j])));}}for(auto i=twoSum.begin();i!=twoSum.end();++i){int sub = target - i->first;auto range = twoSum.equal_range(sub);for(auto j=range.first;j!=range.second;++j){pair<int,int> k = i->second;int a = k.first ;int b = k.second;int c = j->second.first;int d = j->second.second;if(a!=c && a!=d && b != c && b!=d){temp[0] = a;temp[1] = b;temp[2] = c;temp[3] = d;sort(temp.begin(),temp.end());result.push_back(temp);}}}sort(result.begin(), result.end());result.erase(unique(result.begin(),result.end()), result.end());return result;}};int _tmain(int argc, _TCHAR* argv[]){vector<int> v1;vector<vector<int>> result;v1.push_back(1);v1.push_back(0);v1.push_back(-1);v1.push_back(0);v1.push_back(-2);v1.push_back(2);//v1.push_back((rand()%200 - 100));v1.push_back((rand()%200 - 100));Solution s;clock_t start,finish;long i = 100000L;double duration;start = clock();result = s.FourSum1(v1,0);finish = clock();duration = (double)(finish - start)/CLOCKS_PER_SEC;cout<< "FourSum1 Runtime:" << duration << endl;start = clock();result = s.FourSum2(v1,0);finish = clock();duration = (double)(finish - start)/CLOCKS_PER_SEC;cout<< "FourSum2 Runtime:" << duration << endl;start = clock();result = s.FourSum3(v1,0);finish = clock();duration = (double)(finish - start)/CLOCKS_PER_SEC;cout<< "FourSum3 Runtime:" << duration << endl;return 0;}

小结：

很值得研究的一道题目，建议自己写一遍，不仅对算法思想提升有帮助，而且在使用stl的学习上，也是有帮助的。根据我测试，就运行时间来说，方法一最快，二其次，三最慢。

另外，方法二实验unordered_map，其中key值相同的话，相同的key的value保存在一个vector桶中。而方法三使用unordered_multimap,当存在key相同的情况，并不是像前者那样存储，而是分开存储。

比如：unordered_map中形式如：   [1] = (-2, [2]((-2, 0),(-2, 0)))  相同的键值都放在一起。不允许重复的键值。

unordered_multimap中 形式如：(-2, (-2, 0)),(-2, (-2, 0))  虽然键值不一样，但是都是分开存放的。允许重复的键值。