leetcode.18--------------4Sum

标签：4sum   算法   acm   leetcode   

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)

1.

class Solution {
public:
	vector<vector<int> > fourSum(vector<int> &num, int target)
	{
		vector<vector<int> > ret;

		if (num.size() == 0) return ret;

		sort(num.begin(), num.end());

		for (size_t a = 0; a < num.size(); ++a)
		{
			if (a != 0 && num[a] == num[a - 1])
			{
				continue;
			}

			for (size_t b = a + 1; b < num.size(); ++b)
			{
				if (b != a + 1 && num[b] == num[b - 1])
				{
					continue;
				}

				size_t c = b + 1;
				size_t d = num.size() - 1;

				while (c < d)
				{
					const int sum = num[a] + num[b] + num[c] + num[d];

					if (sum > target)
					{
						--d;
					}
					else if (sum < target)
					{
						++c;
					}
					else if (c != b + 1 && num[c] == num[c - 1])
					{
						++c;
					}
					else if (d != num.size() - 1 && num[d] == num[d + 1])
					{
						--d;
					}
					else
					{
						vector<int> result;

						result.push_back(num[a]);
						result.push_back(num[b]);
						result.push_back(num[c]);
						result.push_back(num[d]);

						ret.push_back(result);

						++c;
						--d;
					}
				}
			}
		}

		return ret;
	}
};

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());
		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)
						begin++;
					else
						end--;
				}
			}
		}
		set<vector<int>>::iterator it = tmpres.begin();
		for (; it != tmpres.end(); it++)
			res.push_back(*it);
		return res;
	}
};

leetcode.18--------------4Sum

标签：4sum   算法   acm   leetcode   

原文地址：http://blog.csdn.net/chenxun_2010/article/details/43458333