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)

关键在于对于重复的去处，N1，N2，N3，N4 四项每项都和自己的上一次数值比较，重复则取下一个数值，N1，N2，N3，N4之间允许任意项重复。

class Solution {
public:
    vector<vector<int> > fourSum(vector<int> &num, int target)
	{
		vector< vector<int> > res;
        sort(num.begin(), num.end());
		for (int n4 = num.size() - 1; n4 > 2; n4--)
		{
			if (n4 != num.size() - 1 && num[n4] == num[n4+1])
				continue;
			for (int n3 = n4 - 1; n3 > 1; n3--)
			{
				if (n3 != n4 - 1 && num[n3] == num[n3+1])
					continue;
				int tar = target - num[n3] - num[n4];
				for (int n2 = n3 - 1, n1 = 0; n2 > n1;)
				{
					if(n1 > 0 && num[n1] == num[n1-1]|| (tar > num[n2] + num[n1]))
						n1++;
					else if (n2 < n2 - 1 && num[n2] == num[n2+1]  || (tar < num[n2] + num[n1]))
						n2--;
					else 
					{
						vector<int> tmp;
						tmp.push_back(num[n1]);
						tmp.push_back(num[n2]);
						tmp.push_back(num[n3]);
						tmp.push_back(num[n4]);
						res.push_back(tmp);
						n2--;n1++;
					}
				}
			}
		}
		return res;
    }
};