技術標籤：LeetCode300-javaleetcode

文章目錄

• 題目地址
• 題目描述
• 思路
• 題解

題目地址

中文：https://leetcode-cn.com/problems/count-and-say/
英文：https://leetcode.com/problems/count-and-say/

題目描述

The count-and-say sequence is a sequence of digit strings defined by the recursive formula:

• countAndSay(1) = “1”
• countAndSay(n) is the way you would “say” the digit string from countAndSay(n-1), which is then converted into a different digit string.
  
  To determine how you “say” a digit string, split it into the minimal number of groups so that each group is a contiguous section all of the same character. Then for each group, say the number of characters, then say the character. To convert the saying into a digit string, replace the counts with a number and concatenate every saying.

For example, the saying and conversion for digit string “3322251”:

Given a positive integer n, return the nth term of the count-and-say sequence.

Example 1:

Input: n = 1
Output: "1"
Explanation: This is the base case.

Example 2:

Input: n = 4
Output: "1211"
Explanation:
countAndSay
 
(1) = "1"
countAndSay(2) = say "1" = one 1 = "11"
countAndSay(3) = say "11" = two 1's = "21"
countAndSay(4) = say "21" = one 2 + one 1 = "12" + "11" = "1211"

Constraints:

1 <= n <= 30

思路

題目的定義就是遞迴定義的，所以用遞迴解題~

題解

class Solution {
    public String countAndSay(int n) {
        if(n==1) return "1";
        else {
            String s = countAndSay(n-1);
            String str = "";
            int len = s.length();
            int i = len-1;
            while(i>=0){
                int cnt = 0;
                char a = s.charAt(i);
                while(i>=0&&s.charAt(i)==a){
                    cnt++;
                    i--;
                }
                str =cnt+""+a+str;
                cnt = 0;
            }
            return str;
        }
    }
}