LeetCode18:4Sum

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)

#include<iostream>
#include<algorithm>
#include<vector>
#include<set>
using namespace std;

//先对数组进行从小到大的排序；
//确定四个数中的两个，再遍历剩余数组中其他的数，找到另外两个
class Solution{
public:
	vector<vector<int> > fourSum(vector<int> &num, int target) {
		// Note: The Solution object is instantiated only once.
		vector<vector<int>> res;
		int numlen = num.size();
		if (num.size() < 4)    return res;

		sort(num.begin(), num.end()); //先对数组从小到大进行排序
		set<vector<int>> tmpres;
		for (int i = 0; i < numlen; i++)
		{
			for (int j = i + 1; j < numlen; j++)
			{
				int begin = j + 1;
				int end = numlen - 1;
				while (begin < end)
				{
					int sum = num[i] + num[j] + num[begin] + num[end];
					if (sum == target)
					{
						vector<int> tmp;
						tmp.push_back(num[i]);
						tmp.push_back(num[j]);
						tmp.push_back(num[begin]);
						tmp.push_back(num[end]);
						tmpres.insert(tmp);
						begin++;
						end--;
					}
					else if (sum < target)  //sum太小，begin向后移动 
						begin++;
					else            //sum太大，end向前移动  
						end--;
				}
			}
		}
		set<vector<int>>::iterator it = tmpres.begin();
		for (; it != tmpres.end(); it++)
			res.push_back(*it);
		return res;
	}
};

int main()
{
	vector<int> num1= { 1, 0 ,-1, 0 ,- 2 ,2 };
	int target = 0;
	Solution sol1;
	vector<vector<int>> ans;
	ans = sol1.fourSum(num1, target);
	for (int i = 0; i < ans.size(); ++i)
	{
		vector<int> s1 = ans[i];
		cout << '(';
		for (int j=0; j < s1.size(); ++j)
		{
			cout << s1[j] << ',';
		}
		cout << ')' << endl;
	}
	system("pause");
	return 0;
}