[LeetCode]4Sum

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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)

class Solution {
public:
    vector<vector<int> > fourSum(vector<int> &num, int target) {
     std::vector< std::vector<int> > ans;
	if (num.size() < 4)
	{
		return ans;
	}
	sort(num.begin(), num.end());

	//排序后,从左到右扫描;中间两个数一个从left+1,一个从right-1开始
	for (int left = 0; left < num.size() - 3;)
	{
		for (int right = num.size() - 1; right > left + 2;)
		{
			int ml = left + 1; 
			int mr = right - 1;
			while (mr > ml)
			{
				int temp = num[left] + num[ml] + num[mr] + num[right];
				if (temp > target)
				{
					mr--;
					while(num[mr] == num[mr + 1])
					{
						mr--;
					}
				}
				else if (temp < target)
				{
					ml++;
					while(num[ml] == num[ml - 1])
					{
						ml++;
					}
				}
				else
				{
					std::vector<int> temp;
					temp.push_back(num[left]);
					temp.push_back(num[ml]);
					temp.push_back(num[mr]);
					temp.push_back(num[right]);
					ans.push_back(temp);

					mr--;
					ml++;
					while(ml < mr && num[mr] == num[mr + 1])
					{
						mr--;
					}
					while(ml < mr && num[ml] == num[ml - 1])
					{
						ml++;
					}

				}
			}
			right --;
			while(left < right && num[right] == num[right + 1])
			{
				right--;
			}
		}
		left++;
		while(num[left] == num[left - 1])
		{
			left++;
		}
	}
	return ans;
    }
};

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[LeetCode]4Sum

标签：des   style   blog   class   code   c   

原文地址：http://blog.csdn.net/jet_yingjia/article/details/26010511