題目描述

題目解答

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++
    //    這裏要特別註意~返回任意重復的一個，賦值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) {
        if(numbers==null || length<=0){
            return false;
        }
        
        // 判斷數組是否合法（每個數都在0~n-1之間）
        for(int i=0;i<length;i++){
            if(numbers[i]<0 || numbers[i]>length-1){
                return false;
            }
        }
        
        
 
//找出重復數字
        for(int i=0;i<length;i++){
            while(i!=numbers[i]){
                if(numbers[i]==numbers[numbers[i]]){
                    duplication[0]=numbers[i];
                    return true;
                }
                
                int temp=numbers[i];
                numbers[i]=numbers[temp]; //註意，寫成numbers[i]=numbers[numbers[i]]就錯了
                numbers[temp]=temp;
            }
        }
        
        return false;
    }
}

50.數組中重復的數字