【Leetcode】Count and Say

【题目】

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.
【代码】

class Solution {
public:
string countAndSay(int n) {
if(n<=0) return "";
string str="1";         //str用来保存前一个状态的字符串
while(--n){
int count=1;
string newStr="";
for(int i=0;i<str.length();++i){
if((i<str.length()-1)&&(str[i+1]==str[i])) ++count;
else{
char cTemp=(char)(count+'0');
newStr+=cTemp;
newStr+=str[i];
count=1;
}
}
str=newStr;
}
return str;
}
};

【总结】

1.思路：题目的意思是数字符个数然后说出来，每个字符的表达格式是：字符个数+字符本身。对于一个字符串，顺序扫描，统计个数，生成新的表达字符串。