38. 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.
My Thought
理解题目理解了半天。。。
好吧,大致意思是,第一个数字是“1”(字符串的形式)。从第二个数字开始,字符串按照前一个字符串的读法决定。比如,前一个是“1”,那么就是1个“1”,于是第二个字符串是“11”,依次类推。
一个递归解决问题。
伪代码:
PROCEDURE countAndSay(n)
if n = 1
return "1"
else
// 获取前一个字符串
preString = countAndSay(n-1)
// 按规则读字符串,作为返回串、
return read(preString)
Code(C++ 0ms)
class Solution {
public:
string countAndSay(int n) {
if(n==1)
return "1";
string pre = countAndSay(n-1);
// read rule
pre+='#';
string ret="",temp="";
int num = 1;
char ch=pre[0];
for(int i=1;i<pre.size();++i){
if(pre[i]!=ch){
temp+=('0'+num);
temp+=ch;
ret.append(temp);
temp="";
ch=pre[i];
num=0;
}
num+=1;
}
return ret;
}
};
最后0ms ACC 镇啊,激动(虽然很ez23333
题解在我的github上会第一时间更新,有兴趣的可以star一下