leetcode 18 4Sum

一、题目描述

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] ]

二、思路

目的数组 S 中找出四个元素使之和为 target，与 3sum 类似，只是外面变为了二重循环。

注意去重

三、代码

class Solution {
public:
	vector<vector <int> > fourSum(vector<int>& nums, int target) {
		int i, j;
		vector<vector <int>> res;
		if (nums.size() < 4)
			return res;
		sort(nums.begin(), nums.end());
		if (nums.size() == 4)
		{
			if (nums[0] + nums[1] + nums[2] + nums[3] == target)
			{
				vector<int> ini;
				ini.push_back(nums[0]);
				ini.push_back(nums[1]);
				ini.push_back(nums[2]);
				ini.push_back(nums[3]);
				res.push_back(ini);
			}
			else
			{
			}
			return res;
		}
		for (i = 0; i < nums.size() - 3; i++)
		{
			for (j = i + 1; j < nums.size() - 2; j++)
			{
				int k = j + 1;
				int l = nums.size() - 1;
				while (k < l)
				{
					if (nums[i] + nums[j] + nums[k] + nums[l] > target)
					{
						l--;
					}
					else if (nums[i] + nums[j] + nums[k] + nums[l] < target)
					{
						k++;
					}
					else if (nums[i] + nums[j] + nums[k] + nums[l] == target)
					{
						vector<int> ini;
						ini.push_back(nums[i]);
						ini.push_back(nums[j]);
						ini.push_back(nums[k]);
						ini.push_back(nums[l]);
						res.push_back(ini);
						k++;
						l--;
						while (nums[k - 1] == nums[k])
							k++;
						while (nums[l] == nums[l + 1])
							l--;
					}
				}
				while (nums[j] == nums[j + 1])
					j++;
			}
			while (nums[i] == nums[i + 1])
				i++;
		}
		return res;
	}
};