Problem Solving (conflictphonenumber--conflicting phone numbers)

The topic describes the mobile launch of the Xiaonei Cornet and family network Cornet is very convenient, but in a new phone but there is an embarrassing bug. For example, when the following number is included in the Address book:
1. Xiao Wang: 600
2. Xiao Li: 467654
3. Xiao Zhang: 600010
When you enter 600, the phone will automatically call the king, so you will never be able to call Xiao Zhang. Now, there are a lot of phones that have this problem,<span>nowcoder</span> want to find a way to judge whether the numbers in each number book are in this conflict. Input Description:

Enter more than one set of data.

The first row of each set of data is an integer n, (1≤n≤10000).

Immediately followed by n-line phone numbers, the number is only a number, the length of not more than 11.

Output Description:

One row for each set of inputs: if there are conflicting numbers in the phone book, output "Yes", otherwise output "No".

Example 1 input replication

3
911
97625999
91125426
5
113
12340
123440
12345
98346

Output replication

Yes
No

The code is as follows:

1  PackageCom.yzh.xuexi;2 3 Importjava.util.Arrays;4 ImportJava.util.Scanner;5 6  Public classConflictphonenumber {7 8      Public Static voidMain (string[] args) {9Scanner Scanner =NewScanner (system.in);Ten          while(Scanner.hasnext ()) { One             intn=integer.valueof (Scanner.nextline ()); AString[] Arr=NewString[n]; -              for(inti = 0; i < arr.length; i++) { -arr[i]=scanner.nextline (); the             } - System.out.println (Conflictphonenumber (arr)); -         } - scanner.close (); +  -     } +      A     //Can also be solved using a dictionary tree at     Private StaticString conflictphonenumber (string[] arr) { -Arrays.sort (arr);//use the arrays built-in sort () function to sort the strings by default rules from small to large (you can also implement n㏒n complexity sorting methods such as quick, Hill, merge, and heap sorting to reduce the complexity of the entire algorithm) -         //after a string is sorted from small to large, only the adjacent two strings may have a second string beginning with the first string of relationships -          for(intI=1;i<arr.length; i++){ -             if(Arr[i].indexof (arr[i-1]) ==0) { -                 return"YES"; in             } -         } to         return"NO"; +     } -  the}

Problem Solving (conflictphonenumber--conflicting phone numbers)