本文共 2250 字，大约阅读时间需要 7 分钟。

    
       
           
   
        
    
Relative Sort Arrays Problem
        
    
Given two arrays, arr1 and arr2, where each element in arr2 is unique and appears in arr1, we need to sort arr1 such that the relative order of elements matches arr2. Elements not present in arr2 should be placed at the end in ascending order.
                
    
            
     
Example:
            
     
Input: arr1 = [2,3,1,3,2,4,6,7,9,2,19], arr2 = [2,1,4,3,9,6]
            
     
Output: [2,2,2,1,4,3,3,9,6,7,19]
        
    
                
    
            
     
Approach
            
     
To solve this problem, we can follow these steps:
            
     
                
      
Mapping Elements: First, create a mapping of each element in arr2 to its position. This helps in determining the required order of elements in arr1.
                
      
Sorting arr1: Sort the elements in arr1 based on their positions in arr2. Elements not present in arr2 are placed at the end in ascending order.
            
     
                        
     
Implementation Steps
            
     
In code, we can implement this using a hash table (unordered_map) to store the positions of elements in arr2. Then, sort arr1 using a custom comparator that checks the positions from arr2. Finally, append any elements from arr1 that are not in arr2, sorted in ascending order.
        
    
                
    
            
     
Code Example
            
     
class Solution {public:    vector
       
         relativeSortArray(vector
        
         & arr1, vector
         
          & arr2) {        unordered_map
          
            positionMap;                // Populate the position map        for (int i = 0; i < arr2.size(); ++i) {            positionMap[arr2[i]] = i;        }                // Sort arr1 based on the position map        sort(arr1.begin(), arr1.end(), [&positionMap](int a, int b) {            return positionMap[a] < positionMap[b];        });                // Append elements from arr1 not present in arr2, sorted        vector
           
             extra; for (int num : arr1) { if (positionMap.find(num) == positionMap.end()) { extra.push_back(num); } } sort(extra.begin(), extra.end()); arr1.insert(arr1.end(), extra.begin(), extra.end()); return arr1; }
           
          
         
        
       
        
    
    
   

转载地址：http://ephdz.baihongyu.com/