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[lincode easy]Count and Say

时间:2015-11-14 06:25:12      阅读:266      评论:0      收藏:0      [点我收藏+]

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Count and Say

 

The count-and-say sequence is the sequence of integers beginning as follows:

1, 11, 21, 1211, 111221, ...

1 is read off as "one 1" or 11.

11 is read off as "two 1s" or 21.

21 is read off as "one 2, then one 1" or 1211.

Given an integer n, generate the nth sequence.

 
Example

Given n = 5, return "111221".

Note

The sequence of integers will be represented as a string.

 

--------------------------------------------------------------------

1\ create a function to read the previous string.

2\ when we read the string ,we need to remeber two variables, one to record the count of the same num,one to record the num itself.

3\ keep in mind that the string in java or C# cannot be changed, so we need to use StringBuilder to save the new string.

 

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public class Solution {
    /**
     * @param n the nth
     * @return the nth sequence
     */
    public String countAndSay(int n) {
        // Write your code here
        if(n<1) return "";
        if(n==1) return "1";
        String str1="1";
        String strN="";
        for(int i=1;i<n;i++)
        {
            strN=readStr(str1);
            str1=strN;
        }
        return strN;
    }
    
    public String readStr(String str)
    {
        StringBuffer strNext=new StringBuffer();
        char num=str.charAt(0);
        int count=0;
        for(int i=0;i<str.length();i++)
        {
            if(str.charAt(i)==num)
            {
                count++;
            }
            else
            {
                strNext.append(count);
                strNext.append(num);
                num=str.charAt(i);
                count=1;
            }
        }
        strNext.append(count);
        strNext.append(num);
        return strNext.toString();
    }
}

 


 

[lincode easy]Count and Say

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原文地址:http://www.cnblogs.com/kittyamin/p/4963749.html

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