【LeetCode】38 - Count and Say

最新推荐文章于 2021-03-17 11:00:30 发布

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最新推荐文章于 2021-03-17 11:00:30 发布

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原文链接：http://www.cnblogs.com/irun/p/4716797.html

本文详细介绍了如何实现计数与说法序列算法，从基本概念出发，逐步深入到代码实现细节，包括算法逻辑、变量定义、循环结构等，通过实例演示了如何将当前字符串转换为下一个字符串的过程。

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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.

Solution：下一个字符串= 当前字符串：∑(字符个数+字符)，例如1211，下一个字符串=count(1)+'1'+count(2)+'2'+count(1)+'1'=111221

 1 class Solution {
 2 public:
 3     string countAndSay(int n) {
 4         string ret="1";
 5         if(n==1)return ret;
 6         while(--n){
 7             string next;
 8             char last=ret[0];
 9             int i=1,count=1;
10             while(i<ret.size())
11             {
12                 while(ret[i]==last&&i<ret.size()){
13                     i++;
14                     count++;
15                 }
16                 if(i<ret.size()&&ret[i]!=last){
17                     next+=to_string(count);
18                     next+=last;
19                     
20                     count=1;
21                     last=ret[i];
22                     i++;
23                 }
24             }
25             next+=to_string(count);
26             next+=last;
27             ret=next;
28         }
29         return ret;
30     }
31 };