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题目：
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.
代码：
class Solution { public: string countAndSay(int n) { string tmp1 = "1"; string tmp2 = ""; for ( size_t i = 1; i < n; ++i ) { int digit_count = 1; for ( size_t j = 1; j < tmp1.size(); ++j ) { if ( tmp1[j]==tmp1[j-1] ) { ++digit_count; } else { tmp2 += digit_count+‘0‘; tmp2 += tmp1[j-1]; digit_count = 1; } } tmp2 += digit_count+‘0‘; tmp2 += tmp1[tmp1.size()-1]; tmp1 = tmp2; tmp2 = ""; } return tmp1; } };
tips：
这个题意不太好理解。
简单说就是：第n组字符串是第n-1组字符串的读法；读法的准则就是‘连续出现个数+数字’。
其余的就是处理一下边界case。
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原文地址：http://www.cnblogs.com/xbf9xbf/p/4495332.html
