【LeetCode】5.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.

自己只想到了O(n²)的算法，遍历每个字符，然后从中间向两旁扩，寻找最长子串。显示的结果一直是pending...

http://www.felix021.com/blog/read.php?2040有O(n)的算法

public class Solution {

    public String longestPalindrome(String s) {

        // Start typing your Java solution below

        // DO NOT write main() function

        StringBuilder sb = new StringBuilder();

        sb.append('#');

        for (int i = 0; i < s.length(); i++)

        {

            sb.append(s.charAt(i));

            sb.append('#');

        }

         

        String ss = sb.toString();

        int[] p = new int[ss.length()];

        int mx = 0;

        int id = 0;

        for (int i = 0; i < ss.length(); i++)

        {

            p[i] = mx > i ? Math.min(p[id * 2 -i], mx - i) : 1;

            while (i+p[i] < ss.length() && i - p[i] >=0)

            {

                if (ss.charAt(i + p[i]) == ss.charAt(i-p[i]))

                {

                    p[i]++;

                }

                else

                {

                    break;

                }

            }

            if (i + p[i] > mx)

            {

                mx = i + p[i];

                id = i;

            }

        }

         

        // process result

        int maxIndex = 0;

        int maxLen = 0;

        for (int i = 0; i < ss.length(); i++)

        {

            if( p[i] > maxLen)

            {

                maxLen = p[i];

                maxIndex = i;

            }

        }

        StringBuilder r = new StringBuilder();

        for (int i = maxIndex - p[maxIndex] + 1; i <= maxIndex + p[maxIndex] - 1; i++)

        {

            if(ss.charAt(i) != '#')

            {

                r.append(ss.charAt(i));

            }

        }

        return r.toString();

    }

}