18. 4Sum[M]四数之和

题目

Given an array nums of n integers and an integer target, are there elements a, b, c and d in nums 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.
Example:
Given array nums = [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]
]

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思路

本题思路很简单，有了前面3Sum的基础，这里只要将 target-d，然后就是3Sum的解题方法。

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C++

 vector<vector<int>> fourSum(vector<int>& nums, int target) {
        
        vector<vector<int> > result;
        if(nums.size() < 4)
            return result;
        
        sort(nums.begin(),nums.end());
        
        int pMid = 0;
        int pEnd = 0;
        for(int i = 0;i<nums.size() - 3; i++){
            
            //转化成3Sum问题
            int subTarget=target - nums[i];
            if(i > 0 && nums[i] == nums[i-1])
                continue;
            
            for(int j = i + 1;j<nums.size()-2;j++){
                
                int subTarget2 = subTarget - nums[j];
                pMid = j + 1;
                pEnd = nums.size() - 1;
                if(j > i + 1 && nums[j] == nums[j-1])
                    continue;
                
                while(pMid < pEnd){
                    int sum1 = nums[pMid] + nums[pEnd];
                    
                    if(sum1 < subTarget2){
                        pMid ++;
                    }
                    else if(sum1 > subTarget2){
                        pEnd --;
                    }
                    else{
                        vector<int> tempVec(4,0);
                        tempVec[0] = nums[i];
                        tempVec[1] = nums[j];
                        tempVec[2] = nums[pMid];
                        tempVec[3] = nums[pEnd];
                        
                        result.push_back(tempVec);
                        
                        while(pMid < pEnd && nums[pMid] == nums[pMid+1]) //去除重复元素
                            pMid ++;
                        while(pMid < pEnd && nums[pEnd] == nums[pEnd -1])
                            pEnd --;
                        pMid ++;
                        pEnd --;
                    }
                }
            }
        }
        return result;
    }

Python

转载于:https://www.cnblogs.com/Jessey-Ge/p/10993506.html