leetCode 38.Count and Say (计数和发言) 解题思路和方法

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.

Note: The sequence of integers will be represented as a string.

 

思路：题意实在太难理解了，尤其是英文又不好，只能参看下别人的资料，理解下规则。终于理解，题意是n=1时输出字符串1；n=2时，数上次字符串中的数值个数，因为上次字符串有1个1，所以输出11；n=3时，由于上次字符是11，有2个1，所以输出21；n=4时，由于上次字符串是21，有1个2和1个1，所以输出1211。依次类推，写个countAndSay(n)函数返回字符串。

题意理解之后就好办了，是典型的递归问题，其代码很简单，如下：

 

public class Solution {
    public String countAndSay(int n) {
        if(n == 1){
            return 1;
        }
        //递归调用，然后对字符串处理
        String str = countAndSay(n-1) + *;//为了str末尾的标记，方便循环读数
        char[] c = str.toCharArray();
        int count = 1;
        String s = ;
        for(int i = 0; i < c.length - 1;i++){
        	if(c[i] == c[i+1]){
        		count++;//计数增加
        	}else{
        		s = s + count + c[i];//上面的*标记这里方便统一处理
        		count = 1;//初始化
        	}
        }
        return s;
    }
}