Leetcode: 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.

For example:

Given "aacecaaa", return "aaacecaaa".

Given "abcd", return "dcbabcd".

本题最简单的思路是从后往前判断子串是否是回文串，然后把后面的弄到前面即可，不过这样仅仅能擦边通过测试

Brute Force: O(N^2), space O(1)

public class Solution {
    public String shortestPalindrome(String s) {
        if (s==null || s.length()==0) return "";
        char[] arr = s.toCharArray();
        StringBuffer res = new StringBuffer();
        for (int i=arr.length-1; i>=0; i--) {
            if (isPalin(arr, 0, i)) {
                res = new StringBuffer(s.substring(i+1));
                res = res.reverse();
                break;
            }
        }
        res.append(s);
        return res.toString();
    }
    
    public boolean isPalin(char[] arr, int l, int r) {
        while (l < r) {
            if (arr[l] != arr[r]) return false;
            l++;
            r--;
        }
        return true;
    }
}

DP: O(N^2), space O(N^2)

public class Solution {
    public String shortestPalindrome(String s) {
        //if (s==null || s.length()==0) return "";
        boolean[][] dict = new boolean[s.length()][s.length()];
        getdict(s, dict);
        int i = 0;
        for (i=s.length()-1; i>=0; i--) {
            if (dict[0][i]) break;
        }
        StringBuffer res = new StringBuffer(s.substring(i+1));
        res = res.reverse();
        res.append(s);
        return res.toString();
    }
    
    public void getdict(String s, boolean[][] dict) {
        for (int i=s.length()-1; i>=0; i--) {
            for (int j=i; j<s.length(); j++) {
                if (s.charAt(i)==s.charAt(j) && (j-i<=2 || dict[i+1][j-1]))
                    dict[i][j] = true;
            }
        }
    }
}

KMP介绍：http://kb.cnblogs.com/page/176818/

实现：http://www.cnblogs.com/tonyluis/p/4474658.html

本题KMP：（未深究）参考：http://blog.csdn.net/pointbreak1/article/details/45931551

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;
    }
}