LeetCode 4Sum

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)

这道题和3Sum的思路是一样的。时间复杂度是O(n^3)，增加了一个外循环判断，程序长了点，总的程序结构也是一样的。

 

#include<iostream>
#include<vector>
#include<string>
#include<algorithm>

using namespace std;

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

		int i = 0, j = 0, k = 0, r = num.size()-1;
		while (i < r-2)
		{
			for (j = i+1; j < r-1;)
			{
				//Notice: don't forget to reset the value of r = num.size()-1;
				for (k = j+1, r = num.size()-1; k < r; )
				{
					int temp = num[i] + num[j] + num[k] + num[r];
					if (temp == target)
					{
						fNum[0] = num[i]; fNum[1] = num[j]; fNum[2] = num[k]; fNum[3] = num[r];
						result.push_back(fNum);
						k++; r--;
						//注意判断重复的时候需要放进判断条件括号里面，因为没有移动的时候不需要判断
						//k和r都不一定每次移动的
						while (k<r && num[k] == num[k-1])
						{
							k++;
						}
						while (k<r && num[r] == num[r+1])
						{
							r--;
						}
					}
					else if (temp < target)
					{
						k++;
						while (k<r && num[k] == num[k-1])
						{
							k++;
						}
					}
					else
					{
						r--;
						while (k<r && num[r] == num[r+1])
						{
							r--;
						}
					}

				}
				j++;
				while (j<r-1 && num[j] == num[j-1])
				{
					j++;
				}
			}
			i++;
			while (i<r-2 && num[i] == num[i-1])
			{
				i++;
			}
		}
		return result;
	}
};
int main() 
{
	int a[] = {2,7,9,5,5,3,8,10,-1,-2,-5,-6,-17,-35};
	int len = sizeof(a)/sizeof(int);
	vector<int> vi(a, a+len);
	cout<<"The array is:\n";
	for (auto x:vi)
		cout<<x<<" ";
	cout<<endl;

	Solution solu;
	vector<vector<int> > vt = solu.fourSum(vi, 10);

	for (auto x:vi)
		cout<<x<<" ";
	cout<<endl;
	for (auto x:vt)
	{
		for (auto y:x)
			cout<<y<<" ";
		cout<<endl;
	}

	system("pause");
	return 0;
}

 

 

 12.13update： 更新使用set避免重复。程序更加简洁。 

class Solution {
public:
	vector<vector<int> > fourSum(vector<int> &num, int target)
	{
		int n = num.size()-1;
		if (n <3) return vector<vector<int> >();
		set<vector<int> > svi;
		sort(num.begin(), num.end());
		vector<int> vi(4);
		for (int i = 0; i < n-2; i++)
		{
			for (int j = i+1; j < n-1; j++)
			{
				for (int k = j+1, n = num.size()-1; k < n;)
				{
					int t = num[i]+num[j]+num[k]+num[n];
					if (t == target)
					{
						vi[0] = num[i]; vi[1] = num[j]; vi[2] = num[k]; vi[3] = num[n];
						svi.insert(vi);
						k++; n--;
					}
					else if (t < target)
						k++;
					else n--;
				}//k
			}//j
		}//for i
		return vector<vector<int> >(svi.begin(), svi.end());
	}
};

 

//2014-1-25
class Solution {
public:
	vector<vector<int> > fourSum(vector<int> &num, int target) 
	{
		sort(num.begin(), num.end());
		vector<int> tmp(4);
		vector<vector<int> > rs;
		//i<d-2; j<d-1可以让程序由原来的444,432ms左右优化到240,232ms
		int i = 0, j = 1, k = 2, d = num.size()-1;

		for (; i < d-2; i++)
		{
			for (j = i+1; j < d-1; j++)
			{
				for (k = j+1, d = num.size()-1; k < d; )
				{
					int sum = num[i]+num[j]+num[k]+num[d];
					if (sum > target) d--;
					else if (sum < target) k++;
					else
					{
						tmp[0] = num[i];
						tmp[1] = num[j];
						tmp[2] = num[k];
						tmp[3] = num[d];
						rs.push_back(tmp);
						k++;
						d--;
						while (k<d && num[k]==num[k-1]) k++;
						while (k<d && num[d]==num[d+1]) d--;
					}
					while (j<d-1 && num[j]==num[j+1]) j++;
				}
				while (i<d-2 && num[i]==num[i+1]) i++;
			}
		}
		return rs;
	}
};