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			GOOGLE挑戰賽練習題3及答案(1000分)
		
    
		
    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) 
    

     
    

     
    

     1public class CursorPosition
     2{
     3
     4        public int getPosition(String ks,int len)
     5        {
     6                int pos = 0;
     7                char c ;
     8                for(int i=0;i<ks.length();i++)
     9                {
    10                        c = ks.charAt(i);
    11
    12                        if(c=='E')
    13                        {
    14                                pos = len;
    15                        }
    16                        else if(c=='H')
    17                        {
    18                                pos = 0;
    19                        }
    20                        else if(c=='L')
    21                        {
    22                                if(pos>0)
    23                                {
    24                                  pos--;
    25                            }
    26                        }
    27                        else if(c=='R')
    28                        {
    29                                if(pos<len)
    30                                {
    31                                  pos++;
    32                            }
    33
    34                        }
    35                }
    36
    37        return pos;
    38        }
    39
    40}
    
	

    Feedback
    
	
			
    
				
    
					
    
						# re: GOOGLE挑戰賽練習題3及答案(1000分)   回復
						  
					
    
					2005-11-29 16:08 by superwu
				
    
				
    public class CursorPosition {

    	public int getPosition(String keystrokes, int N){

    		int current=0;

    		int end=N;

    		int start=0;

    		

    		

    		for(int i=0;i<keystrokes.length();i++){

    			char c=keystrokes.charAt(i);

    			switch(c){

    			case 'L':

    				if(current!=start)current--;break;

    			case 'R':

    				if(current!=end)current++;break;

    			case 'H':

    				current=0;break;

    			case 'E':

    				current=N;break;

    			}

    		}

    	return current;

    	}

    	public static void main(String[] args) {

    		CursorPosition c=new CursorPosition2();

    		System.out.println(c.getPosition("ERLLL",10));

    	}

    

    }

    這道題和你做的幾乎一樣，效率上應該沒有太大差別

    但我的了920多分，我覺的是時間的原因，這道題我做的很快就提交了
    
			
    
		
    
			posted on 2006-02-13 21:38 Vincent.Chen 閱讀(3136) 評論(0)  編輯  收藏  所屬分類: 雜文 
		
    
	
    
	
	


	


    
	    
    




    








    

				

    




    







    
        
	



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