38. Count and Say【leetcode】【C++】
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2022-03-23 18:14:56
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The count-and-say sequence is the sequence of integers with the first five terms as following:
1. 1 2. 11 3. 21 4. 1211 5. 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 where 1 ≤ n ≤ 30, generate the nth term of the count-and-say sequence.
Note: Each term of the sequence of integers will be represented as a string.
Example 1:
Input: 1 Output: "1"
Example 2:
Input: 4 Output: "1211"
这道题首先是要理解题意,其实就是1代表字符串"1",从2开始。从左到右统计上一个字符串的相同字符个数,比如5的话是”111221“,也就是从左到右是:3个‘1’,2个‘2’,1个‘1’,那么6就应该是"312211"。
直接上代码:
class Solution {
public:
string countAndSay(int n) {
string a[40];
a[0] = "1";
int count; //统计个数
for(int i = 0;i < n-1;++i)
{
int len = a[i].size();
count = 1; //最少应该是1个,而不是0
for(int j = 0;j < len;++j)
{
if(a[i][j] == a[i][j+1])
{
count++;
}
else
{
a[i+1] += ('0'+count);
a[i+1] += a[i][j];
count = 1 ; //接下来是另一个字符,重置count值,重新统计另一个字符的个数
}
}
}
return a[n-1];
}
};
提交结果:
来进行测试一下:
#include<iostream>
#include<cstring>
using namespace std;
string countAndsay(int n)
{
string a[40];
a[0] = "1";
int count;
for(int i = 0;i < n-1;++i)
{
int len = a[i].size();
count = 1;
for(int j = 0;j < len;++j)
{
if(a[i][j] == a[i][j+1])
{
count++;
}
else
{
a[i+1] += ('0'+count);
a[i+1] += a[i][j];
count = 1 ;
}
}
}
return a[n-1];
}
int main()
{
int n;
while(cin >> n)
{
cout << countAndsay(n)<<endl;
}
return 0;
}
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