实现如下

第一个函数：找出重复的元素

第二个函数：找出重复的元素，并输出重复的次数

class duplicate:        
    def duplicate_1(self,array):
        d = {}
        res = []
        for i in array:
            d[i] =d.get(i,0)+1
        for j in d.keys():
            if d[j]>1:
                res.append(j)
        return res
    
            
    def duplicate_2(self,array):
        d = {}
        res = {}
        for i in array:
            d[i] =d.get(i,0)+1
        for j in d.keys():
            if d[j]>1:
                res[j] = d[j]
        return res

if __name__ =='__main__':
    test = duplicate()        
    print(test.duplicate_1([2,1,4,6,3,2,6,8,3]))
    print(test.duplicate_2('aaacdrtguopuuuf'))

结果为

runfile('C:/Users/DELL/Desktop/python/duplicate.py', wdir='C:/Users/DELL/Desktop/python')
[2, 6, 3]
{'a': 3, 'u': 4}

public class Solution {
    // Parameters:
    //    numbers:     an array of integers
    //    length:      the length of array numbers
    //    duplication: (Output) the duplicated number in the array number,length of duplication array is 1,so using duplication[0] = ? in implementation;
    //                  Here duplication like pointor in C/C++, duplication[0] equal *duplication in C/C++
    //    这里要特别注意~返回任意重复的一个，赋值duplication[0]
    // Return value:       true if the input is valid, and there are some duplications in the array number
    //                     otherwise false
	public boolean duplicate(int numbers[],int length,int [] duplication) {
	    int hash[]=new int[length]; 
	    for(int i=0;i<length;i++){
	    	if(hash[numbers[i]]==0){
	    		hash[numbers[i]]++;
	    	}
	    	else{
	    		duplication[0]=numbers[i];
	    		return true;
	    	}
	    }
	    return false;
    }
}