将Java进行到底
将Java进行到底
posts - 15,  comments - 66,  trackbacks - 0

Problem Statement

     When editing a single line of text, there are four keys that can be used to move the cursor: end, home, left-arrow and right-arrow. As you would expect, left-arrow and right-arrow move the cursor one character left or one character right, unless the cursor is at the beginning of the line or the end of the line, respectively, in which case the keystrokes do nothing (the cursor does not wrap to the previous or next line). The home key moves the cursor to the beginning of the line, and the end key moves the cursor to the end of the line.

You will be given a int, N, representing the number of character in a line of text. The cursor is always between two adjacent characters, at the beginning of the line, or at the end of the line. It starts before the first character, at position 0. The position after the last character on the line is position N. You should simulate a series of keystrokes and return the final position of the cursor. You will be given a String where characters of the String represent the keystrokes made, in order. 'L' and 'R' represent left and right, while 'H' and 'E' represent home and end.

Definition

    
Class: CursorPosition
Method: getPosition
Parameters: String, int
Returns: int
Method signature: int getPosition(String keystrokes, int N)
(be sure your method is public)
    

Constraints

- keystrokes will be contain between 1 and 50 'L', 'R', 'H', and 'E' characters, inclusive.
- N will be between 1 and 100, inclusive.

Examples

0)
    
"ERLLL"
10
Returns: 7
First, we go to the end of the line at position 10. Then, the right-arrow does nothing because we are already at the end of the line. Finally, three left-arrows brings us to position 7.
1)
    
"EHHEEHLLLLRRRRRR"
2
Returns: 2
All the right-arrows at the end ensure that we end up at the end of the line.
2)
    
"ELLLELLRRRRLRLRLLLRLLLRLLLLRLLRRRL"
10
Returns: 3
3)
    
"RRLEERLLLLRLLRLRRRLRLRLRLRLLLLL"
19
Returns: 12

This problem statement is the exclusive and proprietary property of TopCoder, Inc. Any unauthorized use or reproduction of this information without the prior written consent of TopCoder, Inc. is strictly prohibited. (c)2003, TopCoder, Inc. All rights reserved.

答案:


 1
 2public class CursorPosition {
 3    public int getPosition(String keystrokes, int N){
 4        int position = 0;
 5        String s = "";
 6        for(int i = 0; i < keystrokes.length(); i++){
 7            s = keystrokes.substring(i, i+1);
 8            if("L".equals(s)){
 9                if(position == 0continue;
10                position--;
11            }

12            if("R".equals(s)){
13                if(position == N) continue;
14                position++;
15            }

16            if("H".equals(s)){
17                position = 0;
18            }

19            if("E".equals(s)){
20                position = N;
21            }

22
23        }

24
25        return position;
26
27    }

28    /**
29     * @param args
30     */

31    public static void main(String[] args) {
32        CursorPosition cursorPosition = new CursorPosition();
33        int cursor = cursorPosition.getPosition("ERLLL"10);
34        System.out.println("cursor:" + cursor);
35    }

36
37}

38
posted on 2005-11-27 23:42 风萧萧 阅读(929) 评论(2)  编辑  收藏 所属分类: 杂谈

FeedBack:
# re: GOOGLE挑战赛练习题3及答案
2005-11-29 13:13 | jackj
应该可以做一些优化,如,
输入字符串从后往前查,如果碰到'E'或'H',则以后的就不用执行了。
例如'ERLLLHRL',查到倒数第三个字符是'H',在前面的'ERLLL'都可以忽略。  回复  更多评论
  
# re: GOOGLE挑战赛练习题3及答案
2005-11-29 13:53 | 风萧萧
有道理,说得对。
谢谢jackj.  回复  更多评论
  

只有注册用户登录后才能发表评论。


网站导航:
 

<2005年11月>
303112345
6789101112
13141516171819
20212223242526
27282930123
45678910

常用链接

留言簿(8)

随笔分类

随笔档案

文章分类

文章档案

相册

收藏夹

myfriends

opensource

搜索

  •  

最新评论

阅读排行榜

评论排行榜