leetcode oj s_05 最长回文子串

/**
 * 最长回文子字符串
 * 
 * @author 林夕
 */
public class Solution {

    public static void main(String[] args) {
        // String s =
        // "rgczcpratwyqxaszbuwwcadruayhasynuxnakpmsyhxzlnxmdtsqqlmwnbxvmgvllafrpmlfuqpbhjddmhmbcgmlyeypkfpreddyencsdmgxysctpubvgeedhurvizgqxclhpfrvxggrowaynrtuwvvvwnqlowdihtrdzjffrgoeqivnprdnpvfjuhycpfydjcpfcnkpyujljiesmuxhtizzvwhvpqylvcirwqsmpptyhcqybstsfgjadicwzycswwmpluvzqdvnhkcofptqrzgjqtbvbdxylrylinspncrkxclykccbwridpqckstxdjawvziucrswpsfmisqiozworibeycuarcidbljslwbalcemgymnsxfziattdylrulwrybzztoxhevsdnvvljfzzrgcmagshucoalfiuapgzpqgjjgqsmcvtdsvehewrvtkeqwgmatqdpwlayjcxcavjmgpdyklrjcqvxjqbjucfubgmgpkfdxznkhcejscymuildfnuxwmuklntnyycdcscioimenaeohgpbcpogyifcsatfxeslstkjclauqmywacizyapxlgtcchlxkvygzeucwalhvhbwkvbceqajstxzzppcxoanhyfkgwaelsfdeeviqogjpresnoacegfeejyychabkhszcokdxpaqrprwfdahjqkfptwpeykgumyemgkccynxuvbdpjlrbgqtcqulxodurugofuwzudnhgxdrbbxtrvdnlodyhsifvyspejenpdckevzqrexplpcqtwtxlimfrsjumiygqeemhihcxyngsemcolrnlyhqlbqbcestadoxtrdvcgucntjnfavylip";
        String s = "babcbabcbaccba";
        // System.out.println(isPalindrome(s));
        long sTime = System.currentTimeMillis();
        String t = method4(s);
        // String t = longestPalindrome(s);
        long eTime = System.currentTimeMillis();
        System.out.println(t);
        System.out.println("执行时间为" + (eTime - sTime) + "ms"); // 执行时间为75ms
    }

    /**
     * 非常规情况：多个最长回文子字符串长度相等 ? 最笨的办法 时间复杂度为O(n^3)
     * 本身有n^2个字符串，检查一个子串是否为回文串又需要O(n)的时间
     */
    public static String longestPalindrome(String s) {
        int l = s.length();
        String res = null;
        int max = Integer.MIN_VALUE;
        for (int i = 0; i < l; i++) {
            for (int j = i; j < l; j++) {
                String temp = s.substring(i, j + 1);
                if (isPalindrome(temp)) {
                    if (temp.length() > max) {
                        max = temp.length();
                        res = temp;
                    }
                }
            }
        }
        return res;
    }

    /**
     * 判断是否为回文字符串
     */
    public static boolean isPalindrome(String s) {
        int l = s.length();
        int temp = (int) Math.floor((double) l / 2); // floor为向下取整
        for (int i = 0; i < temp; i++) {
            if (s.charAt(i) != s.charAt(l - 1 - i)) {
                return false;
            }
        }
        return true;
    }

    /**
     * 时间复杂度：O(n^2) 空间复杂度：O(n^2)
     * 算法理解：http://articles.leetcode.com/longest-palindromic-substring-part-i
     */
    public static String method2(String s) {
        int n = s.length();
        int longestBegin = 0;
        int maxLen = 1;
        boolean[][] table = new boolean[1000][1000];
        for (int i = 0; i < 1000; i++) {
            for (int j = 0; j < 1000; j++) {
                table[i][j] = false;
            }
        }

        for (int i = 0; i < 1000; i++) {
            table[i][i] = true;
        }

        for (int i = 0; i < n - 1; i++) {
            if (s.charAt(i) == s.charAt(i + 1)) {
                table[i][i + 1] = true;
                longestBegin = i;
                maxLen = 2;
            }
        }

        for (int len = 3; len <= n; len++) {
            for (int i = 0; i < n - len + 1; i++) {
                int j = i + len - 1;
                if (s.charAt(i) == s.charAt(j) && table[i + 1][j - 1]) {
                    table[i][j] = true;
                    longestBegin = i;
                    maxLen = len;
                }
            }
        }
        return s.substring(longestBegin, longestBegin + maxLen);
    }

    /**
     * 时间复杂度：O(n^2) 有2N-1个Center  空间复杂度：O(1)
     */
    public static String method3(String s) {
        int n = s.length();
        if (n == 0) {
            return "";
        }
        String longest = s.substring(0, 1);
        for (int i = 0; i < n - 1; i++) {
            String p1 = expandAroundCenter(s, i, i);
            if (p1.length() > longest.length()) {
                longest = p1;
            }

            String p2 = expandAroundCenter(s, i, i + 1);
            if (p2.length() > longest.length()) {
                longest = p2;
            }
        }
        return longest;
    }

    /**
     * 辅助函数 以c1,c2为中心扩展字符串
     * 
     * @param s
     * @param c1
     * @param c2
     * @return
     */
    public static String expandAroundCenter(String s, int c1, int c2) {
        int l = c1, r = c2;
        int n = s.length();
        while (l >= 0 && r <= n - 1 && s.charAt(l) == s.charAt(r)) {
            l--;
            r++;
        }
        return s.substring(l + 1, r);
    }

    /**
     * 时间复杂度:O(n) 空间复杂度:O(n) Manacher’s algorithm
     * http://articles.leetcode.com/longest-palindromic-substring-part-ii
     */
    public static String method4(String s) {
        String T = preProcess(s);
        int n = T.length();
        int[] P = new int[n];
        int C = 0, R = 0;
        for (int i = 1; i < n - 1; i++) {
            int i_mirror = 2 * C - i;
            P[i] = (R > i) ? Math.min(R - i, P[i_mirror]) : 0;

            while (T.charAt(i + 1 + P[i]) == T.charAt(i - 1 - P[i]))
                P[i]++;

            // If palindrome centered at i expand past R,
            // adjust center based on expanded palindrome.
            if (i + P[i] > R) {
                C = i;
                R = i + P[i];
            }
        }
        // Find the maximum element in P.
        int maxLen = 0;
        int centerIndex = 0;
        for (int i = 0; i < n - 1; i++) {
            if (P[i] > maxLen) {
                maxLen = P[i];
                centerIndex = i;
            }
        }

        return s.substring((centerIndex - 1 - maxLen) / 2, (centerIndex - 1 + maxLen) / 2);
    }

    private static String preProcess(String s) {
        int n = s.length();
        if (n == 0) {
            return "^$";
        }
        String ret = "^";

        for (int i = 0; i < n; i++)
            ret += "#" + s.substring(i, i + 1);

        ret += "#$";
        return ret;

    }

}