LeetCode_Shortest Palindrome

Shortest Palindrome

Given a string s, you are allowed to convert it to a palindrome by adding characters in front of it. Find and return the shortest palindrome you can find by performing this transformation.
（在字符串前面增加字符使其为回文序列）

Example:

1. KMP算法

首先要找到s的最长回文前缀s1，剩余部分为s2，那么将s2反转和原s串拼接在一起，即可得到要求的回文串。这里用到 KMP 算法的 next 数组的求最长回文前缀s1，设置一个字符串tmp = s1 + s2 + ‘#’ + s2’ + s1’，具体实现过程如下：

class Solution:
    def shortestPalindrome(self, s: str) -> str:
        n = len(s)
        if n <= 1:
            return s
        tmp = s + '#' + s[::-1]
        k = 0
        next = [0 for i in range(len(tmp))]
        for i in range(1, len(tmp)):
            while k > 0 and tmp[i] != tmp[k]:
                k = next[k-1]
            if tmp[i] == tmp[k]:
                k += 1
            next[i] = k
        return s[:next[-1]-1:-1] + s