Middle-题目106：5. Longest Palindromic Substring

题目原文：
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.
题目大意：
求出字符串S的最长回文子串。
题目分析：
使用了discuss的神算法，我根本看不懂……求大神解答……（好像还是个 O(n2) 的算法）大概是每向右增加一个字符，就判断能不能向左延伸之类的。
本题有一个线性算法，叫manacher算法，参见https://en.wikipedia.org/wiki/Longest_palindromic_substring#CITEREFManacher1975
源码：（language：java）
方法一：

public class Solution {
    public String longestPalindrome(String s) {
        char[] ca = s.toCharArray();
        int rs = 0, re = 0;
        int max = 0;
        for(int i = 0; i < ca.length; i++) {
            if(isPalindrome(ca, i - max - 1, i)) {
                rs = i - max - 1; re = i;
                max += 2;
            } else if(isPalindrome(ca, i - max, i)) {
                rs = i - max; re = i;
                max += 1;
            }
        }
        return s.substring(rs, re + 1);
    }

    private boolean isPalindrome(char[] ca, int s, int e) {
        if(s < 0) return false;

        while(s < e) {
            if(ca[s++] != ca[e--]) return false;
        }
        return true;
    }
}

方法二：（manacher算法，改编自wikipaedia上的代码）

public class Solution {
    public String longestPalindrome(String s) {
        if (s==null || s.length()==0)
            return "";
        char[] s2 = addBoundaries(s.toCharArray());
        int[] p = new int[s2.length]; 
        int c = 0, r = 0; // Here the first element in s2 has been processed.
        int m = 0, n = 0; // The walking indices to compare if two elements are the same
        for (int i = 1; i<s2.length; i++) {
            if (i>r) {
                p[i] = 0; m = i-1; n = i+1;
            } else {
                int i2 = c*2-i;
                if (p[i2]<(r-i)) {
                    p[i] = p[i2];
                    m = -1; // This signals bypassing the while loop below. 
                } else {
                    p[i] = r-i;
                    n = r+1; m = i*2-n;
                }
            }
            while (m>=0 && n<s2.length && s2[m]==s2[n]) {
                p[i]++; m--; n++;
            }
            if ((i+p[i])>r) {
                c = i; r = i+p[i];
            }
        }
        int len = 0; c = 0;
        for (int i = 1; i<s2.length; i++) {
            if (len<p[i]) {
                len = p[i]; c = i;
            }
        }
        char[] ss = Arrays.copyOfRange(s2, c-len, c+len+1);
        return String.valueOf(removeBoundaries(ss));
    }
    private  char[] addBoundaries(char[] cs) {
        if (cs==null || cs.length==0)
            return "||".toCharArray();

        char[] cs2 = new char[cs.length*2+1];
        for (int i = 0; i<(cs2.length-1); i = i+2) {
            cs2[i] = '|';
            cs2[i+1] = cs[i/2];
        }
        cs2[cs2.length-1] = '|';
        return cs2;
    }
    private  char[] removeBoundaries(char[] cs) {
        if (cs==null || cs.length<3)
            return "".toCharArray();

        char[] cs2 = new char[(cs.length-1)/2];
        for (int i = 0; i<cs2.length; i++) {
            cs2[i] = cs[i*2+1];
        }
        return cs2;
    }    
}

成绩：
方法一：9ms,beats 96.88%,众数13ms,4.50%
方法二：11ms,beats 92.14%
cmershen的碎碎念：
Manacher, Glenn (1975), “A new linear-time “on-line” algorithm for finding the smallest initial palindrome of a string”, Journal of the ACM 22 (3): 346–351, doi:10.1145/321892.321896.
这篇论文详细论证了Manacher算法，如果无聊可以研究一下。
方法二不如方法一快是因为数据量不够大（S.length<10^3）,且此题个人认为应归为hard。