[leetcode] 1616. Split Two Strings to Make Palindrome

[leetcode] 1616. Split Two Strings to Make Palindrome

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: trueExplaination: 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: trueExplaination: 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 <= 10^5a.length == b.lengtha 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 def check(self,a,b): i=0 j=len(a)-1 while(i bool: return self.check(a,b) or self.check(b,a)

参考文献

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

版权声明：本文内容由网络用户投稿，版权归原作者所有，本站不拥有其著作权，亦不承担相应法律责任。如果您发现本站中有涉嫌抄袭或描述失实的内容，请联系我们jiasou666@gmail.com 处理，核实后本网站将在24小时内删除侵权内容。