[LeetCode]4 Sum

类似3Sum，先排序，这后两重for循环，然后对最后的一个数组设两个指针遍历。这里注意重复的问题，例如第一重如果和前面一个数相同则跳过，因为前面的查找肯定包含了本次的情况。

// 4Sum.cpp : 定义控制台应用程序的入口点。
//

#include "stdafx.h"
#include <vector>
#include <algorithm>
#include <map>
#include <iostream>
using namespace std;
class Solution {
public:
	vector<vector<int> > fourSum(vector<int> &num, int target) {
		sort(num.begin(),num.end());
		vector<vector<int>>res;
		if (num.size() < 4)
			return res;
		for (int i = 0; i < num.size()-3; i++)
		{
			if (i>0 && num[i] == num[i - 1])
				continue;
			for (int j = i + 1; j < num.size() - 2; j++)
			{
				if (j>i + 1 && num[j] == num[j - 1])
					continue;
				int p1 = j + 1,p2=num.size()-1;
				int sum = target - num[i] - num[j];
				while (p1 < p2)
				{
					if (p1>j + 1 && num[p1] == num[p1 - 1])
					{
						p1++;
						continue;
					}
					if (p2<num.size()-1&&num[p2] == num[p2 + 1])
					{
						p2--;
						continue;
					}
					if (p1 >= p2)
					{
						break;
					}
					if (num[p1] + num[p2] == sum)
					{
						
						vector<int>tempV;
						tempV.push_back(num[i]);
						tempV.push_back(num[j]);
						tempV.push_back(num[p1]);
						tempV.push_back(num[p2]);
						res.push_back(tempV);
						p2--;
					}
					else if (num[p1]+num[p2]>sum)
					{
						p2--;
					}
					else{
						p1++;
					}
				}
			}
		}
		return res;
	}
};
int _tmain(int argc, _TCHAR* argv[])
{
	Solution ss;
	vector<int>test;
	test.push_back(1);
	test.push_back(0);
	test.push_back(-1);
	test.push_back(0);
	test.push_back(-2);
	test.push_back(2);
	vector<vector<int> > res = ss.fourSum(test,0);
	for (int i = 0; i < res.size(); i++)
	{
		for (int j = 0; j < res[i].size(); j++)
			cout << res[i][j] << ",";
		cout << endl;
	}
	system("pause");
	return 0;
}