[LeetCode] 214. Shortest Palindrome 最短回文串

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.

Example 1:

"aacecaaa"
"aaacecaaa"

Example 2:

"abcd"
"dcbabcd"

给一个字符串s，允许在它的前面加字符使其变成回文字符串，找出最短的回文字符串。

解法1: 暴力匹配算法brute force，无法通过OJ，需要用比较巧妙的方法来解。

解法2: KMP算法。

解法3: Manacher's算法。

参考：从头到尾彻底理解KMP

Java:

public class Solution {
    public String shortestPalindrome(String s) {
        String r = new StringBuilder(s).reverse().toString();
        String t = s + "#" + r;
        int[] next = new int[t.length()];
        for (int i = 1; i < t.length(); ++i) {
            int j = next[i - 1];
            while (j > 0 && t.charAt(i) != t.charAt(j)) j = next[j - 1];
            j += (t.charAt(i) == t.charAt(j)) ? 1 : 0;
            next[i] = j;
        }
        return r.substring(0, s.length() - next[t.length() - 1]) + s;
    }
}　　

Python: KMP

class Solution(object):
    def shortestPalindrome(self, s):
        def getPrefix(pattern):
            prefix = [-1] * len(pattern)
            j = -1
            for i in xrange(1, len(pattern)):
                while j > -1 and pattern[j+1] != pattern[i]:
                    j = prefix[j]
                if pattern[j+1] == pattern[i]:
                    j += 1
                prefix[i] = j
            return prefix
 
        if not s:
            return s
 
        A = s + s[::-1]
        prefix = getPrefix(A)
        i = prefix[-1]
        while i >= len(s):
            i = prefix[i]
        return s[i+1:][::-1] + s

Python: Manacher's

class Solution(object):
    def shortestPalindrome(self, s):
        def preProcess(s):
            if not s:
                return ['^', '$']
            string = ['^']
            for c in s:
                string +=  ['#', c]
            string += ['#', '$']
            return string
 
        string = preProcess(s)
        palindrome = [0] * len(string)
        center, right = 0, 0
        for i in xrange(1, len(string) - 1):
            i_mirror = 2 * center - i
            if right > i:
                palindrome[i] = min(right - i, palindrome[i_mirror])
            else:
                palindrome[i] = 0
 
            while string[i + 1 + palindrome[i]] == string[i - 1 - palindrome[i]]:
                palindrome[i] += 1
 
            if i + palindrome[i] > right:
                center, right = i, i + palindrome[i]
 
        max_len = 0
        for i in xrange(1, len(string) - 1):
            if i - palindrome[i] == 1:
                max_len = palindrome[i]
        return s[len(s)-1:max_len-1:-1] + s

C++:

class Solution {
public:
    string shortestPalindrome(string s) {
        string r = s;
        reverse(r.begin(), r.end());
        string t = s + "#" + r;
        vector<int> next(t.size(), 0);
        for (int i = 1; i < t.size(); ++i) {
            int j = next[i - 1];
            while (j > 0 && t[i] != t[j]) j = next[j - 1];
            next[i] = (j += t[i] == t[j]);
        }
        return r.substr(0, s.size() - next.back()) + s;
    }
};