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.
十天没做题了，赶快秒个小题练练手。
思路：就是数每种数字出现了几次。
string countAndSay(int n) { string s = "1"; while(--n) //从1开始计数 { string stmp; char c[10]; for(int i = 0; i < s.length(); ) { int count = 0; char cur = s[i]; while(i < s.length() && s[i] == cur) //数当前重复出现的数字 { count++; i++; } sprintf(c, "%d%d", count, cur - ‘0‘); stmp += c; } s = stmp; } return s; }
【leetcode】Count and Say （easy）
原文：http://www.cnblogs.com/dplearning/p/4510500.html