Two Substrings

最新推荐文章于 2021-07-30 21:28:40 发布

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本文介绍了一个简单的算法挑战：在一个给定的字符串中寻找两个非重叠的子串“AB”和“BA”。文章通过示例解释了问题，并提供了一段C++代码实现，用于判断字符串是否包含这两个特定的子串组合。

You are given string s. Your task is to determine if the given string s contains two non-overlapping substrings "AB" and "BA" (the substrings can go in any order).

Input

The only line of input contains a string s of length between 1 and 10⁵ consisting of uppercase Latin letters.

Output

Print "YES" (without the quotes), if string s contains two non-overlapping substrings "AB" and "BA", and "NO" otherwise.

Sample test(s)

input

ABA

output

NO

input

BACFAB

output

YES

input

AXBYBXA

output

NO

Note

In the first sample test, despite the fact that there are substrings "AB" and "BA", their occurrences overlap, so the answer is "NO".

In the second sample test there are the following occurrences of the substrings: BACFAB.

In the third sample test there is no substring "AB" nor substring "BA".

给一个字符串，找AB和BA，但是AB和BA不能重合，比如ABA不算。直接就是模拟，分类，先找AB,再找BA，然后还有一个先找BA，再找AB。再定义一个标兵，最后输出。

#include<cstdio>
#include<cstring>
#include<iostream>
using namespace std;
char c[100005];
int main()
{
int i, j, m, n, leap;
scanf("%s", c);
int len = strlen(c); leap = 0; int leap1 = 0;
for (i = 0; i < len - 1; i++)
{
  if (c[i] == 'A'&&c[i + 1] == 'B'&&!leap)
  {
   leap = 1;
   i++;
  }
  else if (leap == 1 && !leap1&&c[i] == 'B'&&c[i + 1] == 'A')
   leap1 = 1;
  if (leap1 == 1)break;
}
if (leap1 == 1)
  cout << "YES" << endl;
else
{
  leap = leap1 = 0;
  for (i = 0; i < len - 1; i++)
  {
   if (c[i] == 'B'&&c[i + 1] == 'A'&&!leap)
   {
    leap = 1;
    i++;
   }
   else if (leap == 1 && !leap1&&c[i] == 'A'&&c[i + 1] == 'B')
    leap1 = 1;
   if (leap1 == 1)break;
  }
  if (leap1 == 1)
   cout << "YES" << endl;
  else
   cout << "NO" << endl;
}
return 0;
}