Leetcode Count and Say

最新推荐文章于 2019-08-14 16:27:56 发布

转载最新推荐文章于 2019-08-14 16:27:56 发布 · 61 阅读

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原文链接：http://www.cnblogs.com/criseRabbit/p/4302324.html

本文详细解释了如何使用循环和计数逻辑来生成计数与说明序列，从1开始，每次读取前一个数字并描述其组成。

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.

这道题其实不是很难，关键是要耐心，每一个后面的数字都是根据前面的数字推算出来的，用一个for循环来数重复的数字的个数，

 1 package Count.and.Say;
 2 
 3 public class CountAndSay {
 4      public String countAndSay(int n) {
 5         if(n==1){
 6              return "1";
 7          }
 8         String curr="1";
 9         for(int i=1;i<n;i++){
10             int size=curr.length();
11             int currIndi=0;
12             StringBuilder tempBuilder=new StringBuilder();
13             while(currIndi<size){    
14                 int count=1;
15                 int k=currIndi+1;
16                 while(k<size&&curr.charAt(k-1)==curr.charAt(k))
17                 {
18                     count++;
19                     k++;
20                 }
21                 tempBuilder.append(count).append(curr.charAt(currIndi));
22                 currIndi=k;
23             }
24             curr=tempBuilder.toString();
25         }
26         return curr.toString();     
27         }
28      public static void main(String []args){
29          CountAndSay service=new CountAndSay();
30          String s=service.countAndSay(6);
31          System.out.println(s);
32      }
33 }

转载于:https://www.cnblogs.com/criseRabbit/p/4302324.html