[LeetCode] Count and Say, Solution

最新推荐文章于 2025-09-10 15:16:07 发布

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原文链接：http://www.cnblogs.com/codingtmd/archive/2013/01/28/5078913.html

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#c/c++

本文介绍了一种基于字符串操作的计数读数序列生成算法，该算法通过迭代过程生成第n项计数读数序列。文章详细展示了从初始序列1开始如何逐步生成后续序列，并提供了完整的C++代码实现。

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 n ^th sequence.

Note: The sequence of integers will be represented as a string.

» Solve this problem

[Thoughts]
string-operation. The only trick thing is Line11. seq[seq.size()] always '\0'. It will help to save an "if" statement.

[Code]

1:    string countAndSay(int n) {  
2:      // Start typing your C/C++ solution below  
3:      // DO NOT write int main() function  
4:      string seq = "1";  
5:      int it = 1;  
6:      while(it<n)  
7:      {  
8:        stringstream newSeq;  
9:        char last = seq[0];  
10:        int count =0;  
11:        for(int i =0; i<= seq.size();i++)  
12:        {  
13:          if(seq[i] ==last)  
14:          {  
15:            count ++;  
16:            continue;  
17:          }  
18:          else  
19:          {  
20:            newSeq<<count<<last;  
21:            last = seq[i];   
22:            count =1;  
23:          }  
24:        }  
25:        seq = newSeq.str();  
26:        it++;  
27:      }  
28:      return seq;  
29:    }