leetcode18. 4Sum

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]
]

class Solution {
public:
    vector<vector<int> > fourSum(vector<int> &num, int target) {
    
        vector<vector<int> > res;
    
        if (num.empty())
            return res;
    
        std::sort(num.begin(),num.end());
    
        for (int i = 0; i < num.size(); i++) {
        
            int target_3 = target - num[i];
        
            for (int j = i + 1; j < num.size(); j++) {
            
                int target_2 = target_3 - num[j];
            
                int front = j + 1;
                int back = num.size() - 1;
            
                while(front < back) {
                
                    int two_sum = num[front] + num[back];
                
                    if (two_sum < target_2) front++;
                
                    else if (two_sum > target_2) back--;
                
                    else {
                    
                        vector<int> quadruplet(4, 0);
                        quadruplet[0] = num[i];
                        quadruplet[1] = num[j];
                        quadruplet[2] = num[front];
                        quadruplet[3] = num[back];
                        res.push_back(quadruplet);
                    
                        // Processing the duplicates of number 3
                        while (front < back && num[front] == quadruplet[2]) ++front;
                    
                        // Processing the duplicates of number 4
                        while (front < back && num[back] == quadruplet[3]) --back;
                
                    }
                }
                
                // Processing the duplicates of number 2
                while(j + 1 < num.size() && num[j + 1] == num[j]) ++j;
            }
        
            // Processing the duplicates of number 1
            while (i + 1 < num.size() && num[i + 1] == num[i]) ++i;
        
        }
    
        return res;
    
    }
};

无论时四个数还是三个数，最终要害点都是寻求一个左右指针，和Container With Most Water https://mp.csdn.net/postedit/81053675

这道题非常类似

对于排好序的数组，只要按照大了--，小了++这样的查询步骤，对于每一个加数因子都要去重。双指针的运用要灵活，导致我现在看见什么都想dp

class Solu  vector<vector<int> > fourSum(vector<int> &num, int target) {
    
        vector<vector<int> > res;
    
        if (num.empty())
            return res;
    
        std::sort(num.begin(),num.end());
    
        for (int i = 0; i < num.size(); i++) {
        
            int target_3 = target - num[i];
        
            for (int j = i + 1; j < num.size(); j++) {
            
                int target_2 = target_3 - num[j];
            
                int front = j + 1;
                int back = num.size() - 1;
            
                while(front < back) {
                
                    int two_sum = num[front] + num[back];
                
                    if (two_sum < target_2) front++;
                
                    else if (two_sum > target_2) back--;
                
                    else {
                    
                        vector<int> quadruplet(4, 0);
                        quadruplet[0] = num[i];
                        quadruplet[1] = num[j];
                        quadruplet[2] = num[front];
                        quadruplet[3] = num[back];
                        res.push_back(quadruplet);
                    
                        // Processing the duplicates of number 3
                        while (front < back && num[front] == quadruplet[2]) ++front;
                    
                        // Processing the duplicates of number 4
                        while (front < back && num[back] == quadruplet[3]) --back;
                
                    }
                }
                
                // Processing the duplicates of number 2
                while(j + 1 < num.size() && num[j + 1] == num[j]) ++j;
            }
        
            // Processing the duplicates of number 1
            while (i + 1 < num.size() && num[i + 1] == num[i]) ++i;
        
        }
    
        return res;
    
    }
};