q5_最长回文子串

package 动态规划.q5_最长回文子串.f1;
/**
 * o(n^2) 以每个字符为中心计算回文长度
 */
class Solution {
    public String getPalindrome(String s, int index) {
        String rs = "";
        int sLen = s.length();
        int i = index;
        int j = index;
        while (j < sLen) {
            if (s.charAt(j) == s.charAt(index)) {
                rs = rs + s.charAt(j);
                j++;
            } else {
                break;
            }
        }
        i--;
        while (i >= 0 && j < sLen) {
            if (s.charAt(i) == s.charAt(j)) {
                rs = s.charAt(i) + rs;
                rs = rs + s.charAt(i);
                i--;
                j++;
            } else {
                break;
            }
        }
        return rs;
    }
    public String longestPalindrome(String s) {
        int maxLen = -1;
        String rs = "";
        for (int i = 0; i < s.length(); i++) {
            String t = getPalindrome(s, i);
            if (t.length() > maxLen) {
                maxLen = t.length();
                rs = t;
            }
        }
        return rs;
    }
}

package 动态规划.q5_最长回文子串.f2;
/**
 * 动态规划 o(n^2)
 * 转移方程：字符串两边界值相等并且子字符串是回文字符串则该字符串是回文字符串
 * dp数组含义：字符串s从i到j的索引子字符串是否是回文字符串
 */
public class Solution {
    public String longestPalindrome(String s) {
        int len = s.length();
        if (len < 2) {
            return s;
        }
        boolean[][] dp = new boolean[len][len];
        for (int i = 0; i < len; i++) {
            dp[i][i] = true;
        }
        int maxLen = 1;
        int start = 0;
        for (int j = 1; j < len; j++) {
            for (int i = 0; i < j; i++) {
                if (s.charAt(i) == s.charAt(j)) {
                    if (j - i < 3) {
                        dp[i][j] = true;
                    } else {
                        dp[i][j] = dp[i + 1][j - 1];
                    }
                } else {
                    dp[i][j] = false;
                }
                if (dp[i][j]) {
                    int curLen = j - i + 1;
                    if (curLen > maxLen) {
                        maxLen = curLen;
                        start = i;
                    }
                }
            }
        }
        return s.substring(start, start + maxLen);
    }
}