Leetcode-字符串-38

38. Count and Say

题目链接

The count-and-say sequence is the sequence of integers with the first five terms as following:

  1. 1
  2. 11
  3. 21
  4. 1211
  5. 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 term of the count-and-say sequence. Note: Each term of the sequence of integers will be represented as a string.


通俗来说,第2项是11,因为第1项是1,读起来就是1个1,所以第2项的11前一个1表示后一个1的计数,以此类推。

第n项:第n-1项的数字串从左到右读出来。


方法1:2ms beat100%

思路:递归 在n-1项的结果上求解。设置首尾指针来确定前后一致的字符个数。apend(相同字符个数).apend(字符)
public String countAndSay(int n) {
        if (n == 1){
            return "1";
        }
        else {
            StringBuilder sb = new StringBuilder();
            String forward = countAndSay(n - 1);
            char[] words = forward.toCharArray();
            for (int i = 0; i < words.length; i++) {
                int j = i;
                while (j < words.length && words[i] == words[j]){
                    j++;
                }
                sb.append(j - i).append(words[i]);
                i = j - 1;
            }
            return sb.toString();
        }
    }



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转载自blog.csdn.net/gaoruowen1/article/details/80940956
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