找水王

1 import java.io.BufferedReader;

 2 import java.io.FileReader;

 3 import java.io.IOException;

 4

 5

 6 public class FindIdlePeople {

 7     public static void main(String[] args) throws IOException {

 8         find();

 9     }

10    

11     public static void find() throws IOException {

12         BufferedReader br = new BufferedReader(new FileReader("list.txt"));

13         String s,s1 = null,s2 = null,shuiwang = null;

14         int i=1;

15         while((s = br.readLine()) != null) {

16             s1 = s;

17             if(s2 == null) {

18                 s2 = br.readLine();

19             }

20             System.out.println("\n第" + i++ + "次比较");

21             System.out.println("s1:" + s1);

22             System.out.println("s2:" + s2);

23             if(s2 != null) {      //判断是否读到末尾即s2是否读了一个空行

24                 if(s1.compareTo(s2) != 0)   //两个值不同，s2重置，因为s1每次都重置，所以不用管s1

25                     s2 = null;

26                 else                        //两个值相同，s2不重置，将相同的值当作水王记录下来

27                     shuiwang = s2;

28             }else {

29                 shuiwang = s1;

30             }

31             System.out.println("记录id:" + shuiwang);

32         }

33         if(shuiwang == null)

34             shuiwang = s2;

35         System.out.println("\n水王id为:" + shuiwang);

36         br.close();

37     }

38    

39 }

设计思想：本题的设计是采用我们都玩过的游戏“消消乐”的算法，由于”水王“的帖子超过了所有帖子的一半，相邻的两个ID可以采用抵消的方法:不同消去,相同合并，最后剩下的一定是水王的帖子。