Description

You are given two strings a and b of the same length. Choose an index and split both strings at the same index, splitting a into two strings: aprefix and asuffix where a = aprefix + asuffix, and splitting b into two strings: bprefix and bsuffix where b = bprefix + bsuffix. Check if aprefix + bsuffix or bprefix + asuffix forms a palindrome.

When you split a string s into sprefix and ssuffix, either ssuffix or sprefix is allowed to be empty. For example, if s = “abc”, then “” + “abc”, “a” + “bc”, “ab” + “c” , and “abc” + “” are valid splits.

Return true if it is possible to form a palindrome string, otherwise return false.

Notice that x + y denotes the concatenation of strings x and y.

Example 1:

Input: a = "x", b = "y"
Output: true
Explaination: If either a or b are palindromes the answer is true since you can split in the following way:
aprefix = "", asuffix = "x"
bprefix = "", bsuffix = "y"
Then, aprefix + bsuffix = "" + "y" = "y", which is a palindrome.

Example 2:

Input: a = "abdef", b = "fecab"
Output: true

Example 3:

Input: a = "ulacfd", b = "jizalu"
Output: true
Explaination: Split them at index 3:
aprefix = "ula", asuffix = "cfd"
bprefix = "jiz", bsuffix = "alu"
Then, aprefix + bsuffix = "ula" + "alu" = "ulaalu", which is a palindrome.

Example 4:

Input: a = "xbdef", b = "xecab"
Output: false

Constraints:

  • 1 <= a.length, b.length <= 10^5
  • a.length == b.length
  • a and b consist of lowercase English letters

分析

题目的意思是：判断两个字符串做同样的分割之后，分别交换首字符串并分别进行拼接，看其中的一个字符串能够构成回文子串。这道题首先把其中一个字符串逆序，然后判断首部字符串和尾部字符串是否有相等的串，来判断是否有回文子串。好像并不work。
后面看见了别人的思路，首先把其中一个字符串的首部和另一个字符串的尾部相同的字符去掉，然后判断剩下索引区间的字符串是否是回文子串，这个规律估计很难被发现，我也是看了之后才知道可以这么干，神奇。

代码

class Solution:
    def isPalindrome(self,a,i,j):
        while(i<j and a[i]==a[j]):
            i+=1
            j-=1
        return i>=j
        
    def check(self,a,b):
        i=0
        j=len(a)-1
        while(i<j and a[i]==b[j]):
            i+=1
            j-=1
        return self.isPalindrome(a,i,j) or self.isPalindrome(b,i,j)
    
    def checkPalindromeFormation(self, a: str, b: str) -> bool:
      
        return self.check(a,b) or self.check(b,a)

参考文献

​​[LeetCode] C++/Java Greedy O(n) | O(1)​​