4Sum Problem

最新推荐文章于 2022-07-07 09:00:00 发布

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最新推荐文章于 2022-07-07 09:00:00 发布

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本文探讨了4Sum问题的不同解决方案，包括使用3Sum相似思路的O(N^3)方法及生成两两数对进行2Sum的O(N^2)方法，并提供了详细的C++实现代码。

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4SumJan 27 '12

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 â‰¤ b â‰¤ c â‰¤ 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)

方法1：利用和3Sum一样的思路，但是复杂度为O(N^3)，偷懒用之前的代码，堆在一起看起来就很屎，懒得改了。

代码：

class Solution {
public:
    vector<vector<int> > fourSum(vector<int> &num, int target) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        sort(num.begin(),num.end());  
  
        vector<vector<int>> ret4;
    	set<vector<int>> set4;
        //visit each positive number as target and use 2Sum to the left ones  
        int target3;  
        vector<int>::iterator it = num.begin();  
        for(;it!=num.end();){  
			if(it!=num.begin() && *it == *(it-1)) {
				it++;
				continue;
			}
            target3 =target-*it;  
  
            //erase the element  
            it = num.erase(it);  
            //3Sum  
            set<vector<int>> set3 = threeSum(num,target3,target-target3);
			set4.insert(set3.begin(),set3.end());

            //insert the element  
            it = num.insert(it,target-target3);  
            it++;  
        }  
        ret4.insert(ret4.end(),set4.begin(),set4.end());  

        return ret4;  
    }set<vector<int> > threeSum(vector<int> &num,int target3,int target) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function

		vector<vector<int>> ret3;
		//visit each positive number as target and use 2Sum to the left ones
		int target2;
		set<vector<int>> set3;
		vector<int>::iterator it = num.begin();
		for(;it!=num.end();){
			if(it!=num.begin() && *it == *(it-1)) {
				it++;
				continue;
			}
			target2 = target3 -*it;

			//erase the element
			it = num.erase(it);
			//2Sum2
			set<vector<int>> set2 = twoSum2(num,target2,target3,target);
			set3.insert(set2.begin(),set2.end());
			//insert the element
			it = num.insert(it,target3-target2);
			it++;
		}
		return set3;
    }     set<vector<int>> twoSum2(vector<int> &numbers, int target2,int target3,int target) {    
        // Start typing your C/C++ solution below     
        // DO NOT write int main() function     
                 
        //match     
        int i = 0;    
        int j = numbers.size()-1;    
        set<vector<int>> ret2;    
       while(i<j){    
            if(numbers[i]+numbers[j]==target2){   
                vector<int> ret;  
                ret.push_back(numbers[i]);  
                ret.push_back(numbers[j]);  
                ret.push_back(target3-target2); 
				ret.push_back(target);
				sort(ret.begin(),ret.end());
                ret2.insert(ret);  
                i++;  
                j--;  
            }else if(numbers[i]+numbers[j]>target2){  
                j--;  
            }else{  
                i++;  
            }      
        }    
        return ret2;   
    }
};

时间：1800 milli secs

方法2：对num生成两两的数对，然后对数对间进行2Sum，这样空间复杂度为O(N^2)，时间复杂度也为O(N^2)。

代码改天补上