标签：
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.
题意是n=1时输出字符串1；n=2时，数上次字符串中的数值个数，因为上次字符串有1个1，所以输出11；n=3时，由于上次字符是11，有2个1，所 以输出21；n=4时，由于上次字符串是21，有1个2和1个1，所以输出1211。依次类推，写个countAndSay(n)函数返回字符串。
C语言超时，代码如下：
char *countAndSay(int n) { if(n <= 0) return NULL; int count = 1,k = 0;; char *say1 = (char*) malloc(sizeof(char)*100000); char *say2 = (char*) malloc(sizeof(char)*100000); say1[0] = ‘1‘; char last = say1[0]; for(int i = 1; i<n; i++){ k = 0; if(i%2){ for(int j = 0; j<strlen(say1); j++){ if(say1[j] == last){ count++; }else{ say2[k++] = (char)(count+‘0‘); say2[k++] = last; count = 1; last = say1[j]; } } say2[k] = ‘0‘; }else{ for(int j = 0; j<strlen(say2); j++){ if(say2[j] == last){ count++; }else{ say1[k++] = (char)(count+‘0‘); say1[k++] = last; count = 1; last = say2[j]; } } say1[k] = ‘\0‘; } } if(n%2) return say1; return say2; }
C++代码如下：
class Solution { public: string unguarded_convert(const string &say) { stringstream ss; int count = 0; char last = say[0]; for (size_t i = 0; i <= say.size(); ++i) { if (say[i] == last) { ++count; } else { ss << count << last; count = 1; last = say[i]; } } return ss.str(); } string countAndSay(int n) { if (n <= 0) return string(); string say = "1"; for (int i = 1; i < n; ++i) { say = unguarded_convert(say); } return say; } };
标签：
原文地址：http://www.cnblogs.com/zhhc/p/4353968.html
