5. Longest Palindromic Substring

class Solution {
    public String longestPalindrome(String s) {
        // 设f[i][j]为以s[i]开头s[j]结尾的字符串是否为回文字符串
        // f[i][j] = f[i + 1, j - 1] && (s[i] == s[j])
        if (s == null || s.length() == 0) {
            return "";
        }
        if (s.length() == 1) {
            return s;
        }
        int n = s.length();
        boolean[][] f = new boolean[n][n];
        // 初始条件：一字符回文（j == i）
        for (int i = 0; i < n; i++) {
            f[i][i] = true;
        }
        // 初始条件：二字符回文（j == i + 1）
        for (int i = 0, j = 1; j < n; i++, j++) {
            if (s.charAt(i) == s.charAt(j)) {
                f[i][j] = true;
            }
        }
        int longestLen = 0;
        int longestI = 0;
        int longestJ = 0;
        for (int i = n - 2; i >= 0; i--) {
            for (int j = n - 1; j >= 0; j--) {
                // 除去初始条件以外的情况
                if (j != i && j != i + 1) {
                    if (j - 1 >= 0) {
                        f[i][j] = f[i + 1][j - 1] && (s.charAt(i) == s.charAt(j));
                    }
                }
                if (f[i][j]) {
                    int len = j - i + 1;
                    if (len > longestLen) {
                        longestLen = len;
                        longestI = i;
                        longestJ = j;
                    }
                }
            }
        }
        return s.substring(longestI, longestJ + 1);
    }
}