N数之和

矫情吗；* 2022-05-10 01:56 165阅读 0赞

**目录**

两数之和

三数之和

四数之和

--------------------

### 两数之和 ###

[https://leetcode-cn.com/problems/two-sum/description/][https_leetcode-cn.com_problems_two-sum_description]

> 给定一个整数数组和一个目标值，找出数组中和为目标值的**两个**数。

你可以假设每个输入只对应一种答案，且同样的元素不能被重复利用。

**示例:**

>     给定 nums = [2, 7, 11, 15], target = 9
>     
>     因为 nums[0] + nums[1] = 2 + 7 = 9
>     所以返回 [0, 1]

**C++代码实现：**

class Solution {
    public:
        vector<int> twoSum(vector<int>& nums, int target) {
            vector<int> twoSum;
            for(int i=0;i<nums.size();i++){
                for(int j=i+1;j<nums.size();j++){
                    if((nums[i]+nums[j])==target){
                        twoSum.push_back(i);
                        twoSum.push_back(j);
                    }
                }
            }
        return twoSum;
    
        }
    };

### 三数之和 ###

[https://leetcode-cn.com/problems/3sum/description/][https_leetcode-cn.com_problems_3sum_description]

> 给定一个包含 *n* 个整数的数组 `nums`，判断 `nums` 中是否存在三个元素 *a，b，c ，*使得 *a + b + c = *0 ？找出所有满足条件且不重复的三元组。

**注意：**答案中不可以包含重复的三元组。

>     例如, 给定数组 nums = [-1, 0, 1, 2, -1, -4]，
>     
>     满足要求的三元组集合为：
>     [
>       [-1, 0, 1],
>       [-1, -1, 2]
>     ]

**C++代码实现：**

class Solution {
    public:
        vector<vector<int>> threeSum(vector<int>& nums) {
            vector<vector<int> > res;
    
        std::sort(nums.begin(), nums.end());//排序
    
        for (int i = 0; i < nums.size(); i++) {
            
            int target = -nums[i];//当输入的两个数值和为n，n-nums[i];
            int front = i + 1;
            int back = nums.size() - 1;
    
            while (front < back) {
    
                int sum = nums[front] + nums[back];
                
    
                if (sum < target)
                    front++;
    
                else if (sum > target)
                    back--;
    
                else {
                    vector<int> triplet(3, 0);//三个数存储
                    triplet[0] = nums[i];
                    triplet[1] = nums[front];
                    triplet[2] = nums[back];
                    res.push_back(triplet);
                    
    
                    while (front < back && nums[front] == triplet[1]) front++;//第二个数重复处理
                    while (front < back && nums[back] == triplet[2]) back--;//第三个数重复处理
                }
                
            }
    
            while (i + 1 < nums.size() && nums[i + 1] == nums[i]) //第一个数重复处理
                i++;
    
        }
        
        return res;
        }
    };

### 四数之和 ###

[https://leetcode-cn.com/problems/4sum/description/][https_leetcode-cn.com_problems_4sum_description]

> 给定一个包含 *n* 个整数的数组 `nums` 和一个目标值 `target`，判断 `nums` 中是否存在四个元素 *a，**b，c* 和 *d* ，使得 *a* + *b* + *c* + *d* 的值与 `target` 相等？找出所有满足条件且不重复的四元组。

**注意：**

答案中不可以包含重复的四元组。

**示例：**

>     给定数组 nums = [1, 0, -1, 0, -2, 2]，和 target = 0。
>     
>     满足要求的四元组集合为：
>     [
>       [-1,  0, 0, 1],
>       [-2, -1, 1, 2],
>       [-2,  0, 0, 2]
>     ]

**C++代码实现：**

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;
        
        }
    };

[https_leetcode-cn.com_problems_two-sum_description]: https://leetcode-cn.com/problems/two-sum/description/
[https_leetcode-cn.com_problems_3sum_description]: https://leetcode-cn.com/problems/3sum/description/
[https_leetcode-cn.com_problems_4sum_description]: https://leetcode-cn.com/problems/4sum/description/