[LeetCode 214] 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".

Solution:

1. O(n * lgn)

check if substring(0, end) is palindrome or not, end from s.length() to 1. If is palindrome, reverse substring(end, length()) + s

this solution will LSE

public String shortestPalindrome(String s) {
        if(s.length() <=1) return s;
        int right = s.length();
        for(int i = right-1; i>=0; i--) {
            if(isPalindrome(s, i)) {
                StringBuilder sb = new StringBuilder(s.substring(i+1));
                sb.reverse();
                return sb.toString() + s;
            }
        }
        return "";
    }
    public boolean isPalindrome(String s, int right) {
        int left = 0;
        while(left<=right) {
            if(s.charAt(left) != s.charAt(right)) return false;
            right--;
            left++;
        }
        return true;
    }

2. Improve from middle search, O(logn * logn)

public String shortestPalindrome(String s) {
        if(s.length() <=1) return s;
        int center = (s.length() - 1) /2;
        String res = "";
        for(int i = center; i>=0; i--) {
            if(s.charAt(i) == s.charAt(i+1)){
                if((res = isPalindrome(s, i, i+1)) != null) return res;
            }
            
            if((res = isPalindrome(s, i, i)) != null) return res;
        }
        return res;
    }
    public String isPalindrome(String s, int left, int right) {
        int i = 1;
        for(;left-i>=0 && right+i< s.length(); i++) {
            if(s.charAt(left-i) != s.charAt(right+i)) break;
        }
        if(left-i>=0) return null;
        StringBuilder sb = new StringBuilder(s.substring(right+i));
        sb.reverse();
        return sb + s;
    }