# Leetcode-Count and Say

## 题目描述

- 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

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.

## 思路

- Base case: n = 0 print “1”
- for n = 1, look at previous string and write number of times a digit is seen and the digit itself. In this case, digit 1 is seen 1 time in a row… so print “1 1”
- for n = 2, digit 1 is seen two times in a row, so print “2 1”
- for n = 3, digit 2 is seen 1 time and then digit 1 is seen 1 so print “1 2 1 1”
- for n = 4 you will print “1 1 1 2 2 1”
- Consider the numbers as integers for simplicity. e.g. if previous string is “10 1” then the next will be “1 10 1 1” and the next one will be “1 1 1 10 2 1”

## AC代码

```
#include <iostream>
#include <cstdio>
#include <string>
using namespace std;
string countAndSay(int n)
{
string s = "1";
string next;
if(n == 1)
return s;
for(int i = 1; i <n; i++)
{
int j = 0;
while(j < s.size())
{
int k = j + 1;
while(k < s.size() && s[k] == s[j])
k++;
next += (k - j + '0');
next += s[j];
j = k;
}
s = next;
next = "";
}
return s;
}
int main()
{
int n;
#ifndef cout
freopen("in3.in", "r", stdin);
freopen("std3.out","w",stdout);
#endif
while(cin>>n)
cout<<countAndSay(n)<<endl;
#ifndef cout
fclose(stdin);
fclose(stdout);
#endif
return 0;
}
```