LeetCode 题解（91）: Shortest Palindrome

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题目：

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.

For example:

Given "aacecaaa", return "aaacecaaa".

Given "abcd", return "dcbabcd".

题解：

先制作原字符串的对称镜像字符串，如s = "abcd", 镜像a = "abcddcba"。

然后对新字符串a，按KMP算法求Prefix的方法，求Prefix, 得【0， 0， 0， 0， 0， 0， 0， 1】，然后s.length() - prefix[2 * s.length()-1] 即为需要复制到s前的s尾部字符串的个数。

c++版：

class Solution {public:    string shortestPalindrome(string s) {        string result = "";                if(s.length() == 0)            return result;        string a = s + string(s.rbegin(), s.rend());        vector<int> prefix(2 * s.length());        for(int i = 1; i < 2 * s.length(); i++) {            int j = prefix[i-1];            while(j > 0 && a[i] != a[j])                j = prefix[j-1];            if(a[i] == a[j])                j++;            prefix[i] = j;        }        int count = s.length() - equal[2 * s.length()-1];        return string(s.rbegin(), s.rbegin() + count) + s;    }};

Java 版：

public class Solution {    public String shortestPalindrome(String s) {        String result = "";        if(s.length() == 0)            return result;        int[] prefix = new int[s.length() * 2];        String mirror = s + new StringBuilder(s).reverse().toString();        for(int i = 1; i < s.length() * 2; i++) {            int j = prefix[i-1];            while(mirror.charAt(j) != mirror.charAt(i) && j > 0)                j = prefix[j-1];            if(mirror.charAt(i) == mirror.charAt(j))                prefix[i] = j + 1;            else                prefix[i] = 0;        }        int count = s.length() - prefix[s.length() * 2 -1];        result = new StringBuilder(s.substring(s.length()-count, s.length())).reverse().toString() + s;        return result;    }}

Python版：

class Solution:    # @param {string} s    # @return {string}    def shortestPalindrome(self, s):        result = ""        if len(s) == 0:            return result        s1 = s + s[::-1]        pre = [0] * len(s1)        for i in range(1, len(s1)):            j = pre[i-1]            while j > 0 and s1[i] != s1[j]:                j = pre[j-1]            if s1[i] == s1[j]:                pre[i] = j + 1            else:                pre[i] = 0                        count = len(s) - pre[len(s1)-1]        s2 = s[len(s)-1:len(s)-count-1:-1]        return s2 + s

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