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.

首先是暴力解，判断S中每一个子字符串是否是回文。时间复杂度O(n^3)

public String longestPalindrome(String str) {
    int max = 0;
    String longest = null;
    for (int i = 0; i < str.length(); i++) {
        for (int j = i + 1; j < str.length(); j++) {
            String subString = str.substring(i, j + 1);
            if (isPalindormic(subString)) {
                if (subString.length() > max) {
                    max = subString.length();
                    longest = subString;
                }
            }
        }
    }
    return longest;
}

private boolean isPalindormic(String s) {
    for (int i = 0, j = s.length() - 1; i < j; i++, j--) {
        if (s.charAt(i) != s.charAt(j)) {
            return false;
        }
    }
    return true;
}

这毕竟还是一道DP题。 如果dp[i][j]定义的是从i到j位置的回文字符串，那只要第i-1和j+1位置上的字符相等就能推出dp[i-1][j+1]也是回文。（前提是i > j + 1)

public String longestPalindrome(String str) {
    boolean[][] table = new boolean[str.length()][str.length()];
    int max = 0;
    String longest = null;
    for (int i = str.length() - 1; i >= 0; i--) {
        for (int j = i; j < str.length(); j++) {
            if (str.charAt(i) == str.charAt(j) && ((j - i <= 2) || table[i + 1][j - 1])) {
                table[i][j] = true;
                if (max < j - i + 1) {
                    max = j - i + 1;
                    longest = str.substring(i, j + 1);
                }
            }
        }
    }
    return longest;
}