LeetCode Count and say

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.

该题 解题主要是模拟过程 打印结果

class Solution {public:    string countAndSay(int n) {        string seq = "1";int count = 1;string rs;int len=1;while(count != n){for(int j=0;j<seq.size();){string tmp = "s";tmp[0] = seq[j];int pos = seq.find_first_not_of(tmp,j);if(pos==-1)pos = seq.size();len = pos-j;rs += len+'0';rs += seq[j];j = j+len;len = 1;}seq = rs;rs = "";count++;}return seq;    }};

用了 string::find_first_not_of  不过感觉这个算法复杂度很高 我测试30 就很慢 

网上找到了一个也是用了STL的算法 比我这个要精练很多 

（原作连接找不到了 但是我保留了 他的作者信息）

<pre name="code" class="cpp">// LeetCode, Count and Say// @author 连城 (http://weibo.com/lianchengzju)// 时间复杂度 O(n^2)，空间复杂度 O(n)class Solution {public:        string countAndSay(int n) {            string s("1");            while (--n)                s = getNext(s);            return s;        }        string getNext(const string &s) {            stringstream ss;            for (auto i = s.begin(); i != s.end(); ) {                auto j = find_if(i, s.end(), bind1st(not_equal_to<char>(), *i));                ss << distance(i, j) << *i;                i = j;            }            return ss.str();        }};