Leetcode_count-and-say(c++ and python version)

地址：http://oj.leetcode.com/problems/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.
思路：模拟题。
参考代码：

class Solution {
public:
string countAndSay(int n) {
string str;
if(!n)
return str;
str = "1";
int len, cnt;
char ch;
queue<string>q;
for(int i = 1; i<n; ++i)
{
len = str.length();
cnt = 0;
ch = str[0];
for(int j = 0; j<len; ++j)
{
if(str[j]==ch)
++cnt;
else
{
string tmp;
tmp += char(cnt+'0');
tmp += ch;
q.push(tmp);
cnt = 1;
ch = str[j];
}
}
string tmp;
tmp += char(cnt+'0');
tmp += ch;
q.push(tmp);
tmp.clear();
while(!q.empty())
{
tmp+=q.front();
q.pop();
}
str = tmp;
}
return str;
}
};

参考代码二，记录字符和该字符个数，不要忘记最后一次的记录

class Solution {
public:
string countAndSay(int n) {
string newStr, str = "1";
char ch;
int cnt;
for(int i = 1; i<n; ++i) {
newStr = "";
ch = str[0];
cnt = 1;
for(int j = 1; j<str.length(); ++j) {
if(str[j]==ch)
++cnt;
else {
newStr += (cnt+'0');
newStr += ch;
ch = str[j];
cnt = 1;
}
}
newStr += (cnt+'0');
newStr += ch;
str = newStr;
}
return str;
}
};

Python

class Solution:
# @return a string
def countAndSay(self, n):
s = "1"
for i in range(1, n):
ch = s[0]
cnt = 1
news = ""
for j in range(1, len(s)):
if ch == s[j] :
cnt += 1
else :
news += (str(cnt) + ch)
ch = s[j]
cnt = 1
news += (str(cnt) + ch)
s = news
return s