【Count and Say】cpp

时间:2023-03-09 00:27:35
【Count and Say】cpp

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) {
string tmp1 = "";
string tmp2 = "";
for ( size_t i = ; i < n; ++i )
{
int digit_count = ;
for ( size_t j = ; j < tmp1.size(); ++j )
{
if ( tmp1[j]==tmp1[j-] )
{
++digit_count;
}
else
{
tmp2 += digit_count+'';
tmp2 += tmp1[j-];
digit_count = ;
}
}
tmp2 += digit_count+'';
tmp2 += tmp1[tmp1.size()-];
tmp1 = tmp2;
tmp2 = "";
}
return tmp1;
}
};

tips:

这个题意不太好理解。

简单说就是:第n组字符串是第n-1组字符串的读法;读法的准则就是‘连续出现个数+数字’。

其余的就是处理一下边界case。

=======================================

第二次过这道题,对题意理解好了之后,代码一次AC。

class Solution {
public:
string countAndSay(int n) {
if (n<) return "";
string ret = "";
for ( int i=; i<n; ++i )
{
string tmp_ret = "";
int count_same_digit = ;
char curr_digit = ret[];
for ( int j=; j<ret.size(); ++j )
{
if ( ret[j]!=ret[j-] )
{
tmp_ret += string(,count_same_digit+'') + string(,curr_digit);
curr_digit = ret[j];
count_same_digit = ;
}
else
{
count_same_digit++;
}
}
tmp_ret += string(,count_same_digit+'') + string(,curr_digit);
ret = tmp_ret;
}
return ret;
}
};