[LeetCode.18]四数之和

根据:https://blog.csdn.net/weixin_39460182/article/details/80209214

改进一个版本

code:

class Solution
{
public:
	vector<vector<int>> fourSum(vector<int>& nums, int target)
	{
		vector<vector<int>> res;
		sort(nums.begin(), nums.end());
		for (int m = 0; m<nums.size(); m++)
		{
			int tarCopy = target;
			if (m > 0 && nums[m] == nums[m - 1])
				continue;
			tarCopy -= nums[m];
			for (int k = m + 1; k < nums.size(); ++k)    //三数之和的代码
			{
				int tarCopy2 = tarCopy;
				if (k > m + 1 && nums[k] == nums[k - 1])
					continue;
				tarCopy2 -= nums[k];
				int i = k + 1, j = nums.size() - 1;
				while (i < j)
				{
					if (nums[i] + nums[j] == tarCopy2)
					{
						res.push_back({ nums[m], nums[k], nums[i], nums[j] });
						while (i < j && nums[i] == nums[i + 1])
							++i;
						while (i < j && nums[j] == nums[j - 1])
							--j;
						++i;
						--j;
					}
					else if (nums[i] + nums[j] < tarCopy2)
						++i;
					else
						--j;
				}
			}
		}
		return res;
	}
};
leetcode能上82.72%,算不错的了。

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转载自blog.csdn.net/weixin_39460182/article/details/80282467