Leetcode 2190. The number that appears most frequently in the array immediately after the key (yes, once)

I'll give you a subscript from 0 The starting array of integers nums , I'll give you an integer at the same time key , It's in nums There have been .

Statistics stay nums The array is followed by key Different integers that appear later target Number of occurrences of . In other words ,target The number of occurrences of is i Number of ：

• 0 <= i <= n - 2
• nums[i] == key And
• nums[i + 1] == target .

Please return and appear most Times target . The test data guarantees the most frequent occurrence of target Is the only one. .

Example 1：

 Input ：nums = [1,100,200,1,100], key = 1
 Output ：100
 explain ： about  target = 100 , In subscript  1  and  4  There has been  2  Time , And they all follow  key .
 There are no other integers in  key  Followed by , So we go back to  100 .

Example 2：

 Input ：nums = [2,2,2,2,3], key = 2
 Output ：2
 explain ： about  target = 2 , In subscript  1 ,2  and  3  There has been  3  Time , And they all follow  key .
 about  target = 3 , In subscript  4  Have you ever appeared  1  Time , And follow  key .
target = 2  Yes, follow closely  key  The number that appears most frequently after , So we go back to  2 .

Tips ：

• 2 <= nums.length <= 1000
• 1 <= nums[i] <= 1000
• The test data ensure that the answer is unique .

Main idea ： utilize map To store
Code:

class Solution {
    
public:
    typedef pair<int, int> PAIR;  
    
    struct CmpByValue {
      
        bool operator()(const PAIR& lhs, const PAIR& rhs) {
      
            return lhs.second > rhs.second;  
        }  
    };
    int mostFrequent(vector<int>& nums, int key) {
    
        map<int,int>mymap;
        for(int i=0;i<nums.size();i++)
        {
    
            if(nums[i]==key)
            {
    
                if((i+1)<nums.size())
                {
    
                    mymap[nums[i+1]]++;
                }
            }
        }
        
        // hold map Elements in are transferred to vector in  
        vector<PAIR> vec(mymap.begin(), mymap.end());  
        
        // Yes vector Sort 
        sort(vec.begin(), vec.end(), CmpByValue());  
        
        return vec[0].first;
    }
};