Longest Palindromic Substring

Given a string S, find the longest palindromic substring in S. You may assume that the maximum length of S is 1000, and there exists one unique longest palindromic substring.

public class Solution {
    String longestOne = null;
    int lengthOfLongest = 0;
    public String longestPalindrome(String s) {
        // Start typing your Java solution below
        // DO NOT write main() function
        longestOne = null;
        lengthOfLongest = 0;
        if(s.length() == 0){
            return longestOne;
        }
        if(s.length() == 1){
            return s;
        }
        for(int i = 0; i < s.length() - 1; i ++){
            if(s.charAt(i) == s.charAt(i+1)){
                if((i == 0 || i == s.length() - 2) && lengthOfLongest < 2){
                        lengthOfLongest = 2;
                        longestOne = s.substring(i,i+2);
                }
                else{getLongestPalindrome(i - 1, i + 2, s);}
            }
        }
        for(int i = 1; i < s.length() - 1; i ++){
            getLongestPalindrome(i - 1, i + 1, s);
        }
        return longestOne;
    }
    
    public void getLongestPalindrome(int left, int right, String s){
        if(left < 0 || right >= s.length()) return;
        int j = right;
        for(int i = left; i > -1 ; i --){
            if(j < s.length() && s.charAt(i) == s.charAt(j)){
                if(j-i+1 >= lengthOfLongest){
                    lengthOfLongest = j-i+1;
                    longestOne = s.substring(i,j+1);
                }
            }
            else{
                return;
            }
            j ++;
        }
    }
}

 第二遍：

 可以将长度为奇数的substring 和 偶数substring 归并到一起。 即使用两倍长度进行循环。 如果是点 则以为中心， 如果是中间，则以两边的点为中心。 减少很多test case。

public class Solution {
    public String longestPalindrome(String s) {
        // Note: The Solution object is instantiated only once and is reused by each test case.
        String longestOne = null;
        int max = 0;
        int len = s.length();
        for(int i = 0; i <= (len - 1)* 2; i ++){
            int left = (i % 2 == 1 ? i / 2 : i / 2 - 1);
            int right = i / 2 + 1;
            while(left > -1 && right < len && s.charAt(left) == s.charAt(right)){
                left --;
                right ++;
            }
            if(right - left - 1 > max){
                max = right - left - 1;
                longestOne = s.substring(left + 1, right);
            }
        }
        return longestOne;
    }
}

 第三遍：

public class Solution {
    public String longestPalindrome(String s) {
        if(s == null || s.length() == 0) return null;
        int len = s.length();
        int max = 0;
        int max_left = 0, max_right = 0;
        for(int i = 0; i < len * 2 - 1; i ++){
            int left  = i / 2;
            int right = (i + 1) / 2;
            while(left > -1 && right < len && s.charAt(left) == s.charAt(right)){
                left --;
                right ++;
            }
            if(right - left > max){
                max = right - left;
                max_left = left + 1;
                max_right = right;
            }
        }
        return s.substring(max_left,max_right);
    }
}