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;
    }
}

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