Leetcode Solution of Longest Palindromic Substring in Java

By X
Wang

Finding the longest palindromic substring is a classic problem of coding interview. In this post, I will summarize 3 different solutions for this problem.

1. Naive Approach

Naively, we can simply examine every substring and check if it is palindromic. The time complexity is O(n^3). If this is submitted to LeetCode onlinejudge, an error message will be returned – “Time Limit Exceeded”. Therefore, this approach is just a start,
we need better algorithm.

public static String longestPalindrome1(String s) {

int maxPalinLength = 0;
String longestPalindrome = null;
int length = s.length();

// check all possible sub strings
for (int i = 0; i < length; i++) {
for (int j = i + 1; j < length; j++) {
int len = j - i;
String curr = s.substring(i, j + 1);
if (isPalindrome(curr)) {
if (len > maxPalinLength) {
longestPalindrome = curr;
maxPalinLength = len;
}
}
}
}

return longestPalindrome;
}

public static boolean isPalindrome(String s) {

for (int i = 0; i < s.length() - 1; i++) {
if (s.charAt(i) != s.charAt(s.length() - 1 - i)) {
return false;
}
}

return true;
}

2. Dynamic Programming

Let s be the input string, i and j are two indices of the string.

Define a 2-dimension array “table” and let table[i][j] denote whether substring from i to j is palindrome.

Start condition:

table[i][i] == 1;
table[i][i+1] == 1  => s.charAt(i) == s.charAt(i+1)

Changing condition:

table[i][j] == 1 => table[i+1][j-1] == 1 && s.charAt(i) == s.charAt(j)

Time O(n^2) Space O(n^2)

public static String longestPalindrome2(String s) {
if (s == null)
return null;

if(s.length() <=1)
return s;

int maxLen = 0;
String longestStr = null;

int length = s.length();

int[][] table = new int[length][length];

//every single letter is palindrome
for (int i = 0; i < length; i++) {
table[i][i] = 1;
}
printTable(table);

//e.g. bcba
//two consecutive same letters are palindrome
for (int i = 0; i <= length - 2; i++) {
if (s.charAt(i) == s.charAt(i + 1)){
table[i][i + 1] = 1;
longestStr = s.substring(i, i + 2);
}
}
printTable(table);
//condition for calculate whole table
for (int l = 3; l <= length; l++) {
for (int i = 0; i <= length-l; i++) {
int j = i + l - 1;
if (s.charAt(i) == s.charAt(j)) {
table[i][j] = table[i + 1][j - 1];
if (table[i][j] == 1 && l > maxLen)
longestStr = s.substring(i, j + 1);
} else {
table[i][j] = 0;
}
printTable(table);
}
}

return longestStr;
}
public static void printTable(int[][] x){
for(int [] y : x){
for(int z: y){
System.out.print(z + " ");
}
System.out.println();
}
System.out.println("------");
}

Given an input, we can use printTable method to examine the table after each iteration. For example, if input string is “dabcba”, the final
matrix would be the following:

1 0 0 0 0 0
0 1 0 0 0 1
0 0 1 0 1 0
0 0 0 1 0 0
0 0 0 0 1 0
0 0 0 0 0 1

From the table, we can clear see that the longest string is in cell table[1][5].

3. Simple Algorithm

Time O(n^2), Space O(1)

public String longestPalindrome(String s) {
if (s.isEmpty()) {
return null;
}

if (s.length() == 1) {
return s;
}

String longest = s.substring(0, 1);
for (int i = 0; i < s.length(); i++) {
// get longest palindrome with center of i
String tmp = helper(s, i, i);
if (tmp.length() > longest.length()) {
longest = tmp;
}

// get longest palindrome with center of i, i+1
tmp = helper(s, i, i + 1);
if (tmp.length() > longest.length()) {
longest = tmp;
}
}

return longest;
}

// Given a center, either one letter or two letter,
// Find longest palindrome
public String helper(String s, int begin, int end) {
while (begin >= 0 && end <= s.length() - 1 && s.charAt(begin) == s.charAt(end)) {
begin--;
end++;
}
return s.substring(begin + 1, end);
}

4. Manacher’s Algorithm

Manacher’s algorithm is much more complicated to figure out, even though it will bring benefit of time complexity of O(n).

Since it is not typical, there is no need to waste time on that.