18. 4Sum

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18. 4Sum

题目

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: 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]
]

解析

class Solution_18 {
public:
	vector<vector<int>> fourSum(vector<int>& nums, int target) {

		vector<vector<int>> vecs;
		
		if (nums.size()<4)
		{
			return vecs;
		}
		sort(nums.begin(), nums.end());
		for (int first = 0; first < nums.size() - 3;first++)
		{
			if (first>0&&nums[first]==nums[first-1])
			{
				continue;
			}

			for (int seconde = first + 1; seconde < nums.size() - 2;seconde++)
			{
				if (seconde>first+1&&nums[seconde]==nums[seconde-1])
				{
					continue;
				}

				int third = seconde + 1;
				int fouth = nums.size() - 1;

				while (third<fouth)
				{
					vector<int> temp = {nums[first],nums[seconde],nums[third],nums[fouth]};
					int sum = accumulate(temp.begin(), temp.end(), 0);
					if (sum==target)
					{
						vecs.push_back(temp);
						third++; fouth--;  //细节问题
						while (third < fouth&&nums[third] == nums[third - 1])
						{
							third++;
						}
						while (third < fouth&&nums[fouth] == nums[fouth + 1])
						{
							fouth--;
						}
	                    // third++; fouth--;放在后面有bug

						//// start++; end--放在后面就是和后面的数比较
						//while (start < end&&nums[start] == nums[start + 1])
						//{
						//	start++;
						//}
						//while (start < end&&nums[end] == nums[end - 1])
						//{
						//	end--;
						//}
						//start++; end--;

						////或者用do() while()循环
						//do{
						//	k++;
						//} while (k < l && ivec[k] == ivec[k - 1]);
						//do{
						//	l--;
						//} while (k < l && ivec[l] == ivec[l + 1]);
					}
					else if(sum<target)
					{
						third++;
					}
					else
					{
						fouth--;
					}
				}
			}
		}
		return vecs;
	}
};

题目来源

• 18. 4Sum
• [C++]LeetCode: 71 4Sum && kSum总结