The count-and-say sequence is the sequence of integers beginning as follows:
1, 11, 21, 1211, 111221, ...
1is read off as"one 1"or11.
11is read off as"two 1s"or21.
21is read off as"one 2, thenone 1"or1211.
Given an integer n, generate the n th sequence.
Note: The sequence of integers will be represented as a string.
题意：返回第n个序列，第i+1个字符串是第i个字符串的读法。参考：Grandyang和JustDoIT的博客。
思路：算法就是对于前一个数，找出相同元素的个数，把个数和该元素存到新的string里。代码需要两个循环，第一个是为找到第n个，第二是为了，根据上一字符串的信息来实现当前的字符串。
1 class Solution { 2 public: 3 string countAndSay(int n) 4 { 5 if(n<1) return NULL; 6 7 string res="1"; 8 for(int i=1;i<n;++i) 9 { 10 string temp; //当前序列 11 res.push_back(‘*‘); 12 int count=0; //重复的个数 13 for(int j=0;j<res.size();++i) 14 { 15 if(j==0) 16 count++; 17 else 18 { 19 if(res[j] !=res[j-1]) 20 { 21 temp.push_back(count+‘0‘); 22 temp.push_back(res[j-1]); 23 count=1; 24 } 25 else 26 ++count; 27 } 28 } 29 res=temp; 30 } 31 return res; 32 } 33 };
原文：http://www.cnblogs.com/love-yh/p/7054922.html