[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;
}