动态规划

最长回文子串
dp

class Solution {
    public String longestPalindrome(String s) {
        int len = s.length();
        int maxLen = 1;
        boolean[][] dp = new boolean[len][len];
        String res = new String("");
        for (int i=0; i<len; i++) {
            for (int j=0;j<len-i;j++) {
                if (i==0) {
                    dp[j][j+i] = true;
                    res = s.substring(j, j+1);
                }
                else if (i==1) {
                    if (s.charAt(j) == s.charAt(j+i)) {
                        dp[j][j+1] = true;
                        res = s.substring(j, j+2);
                    }
                    else {
                        dp[j][j+1] = false;
                    }
                } else {
                    dp[j][j+i] = dp[j+1][j+i-1] && (s.charAt(j) == s.charAt(j+i));
                    res = (dp[j][j+i]) ? s.substring(j, j+i+1) : res;
                }
            }
        }
        return res;
    }
}

马拉车算法
https://www.jianshu.com/p/392172762e55
每个字符都从自己向左右扩展，后面的字符可以根据条件，借助已经访问过的字符，减少扩展次数，将时间复杂度降到O(n)。

class Solution {
    public String longestPalindrome(String s) {
        if (s == null || s.length() <= 1)
            return s;
        String s_temp = "#";
        for (char c : s.toCharArray()) {
            s_temp += c;
            s_temp += '#';
        } 

        int rignthEdge = -1;
        int midIndex = -1;
        int maxLen = -1;
        int initIndex = -1;
        int[] r = new int[s_temp.length()];
        for (int i=0;i<s_temp.length();i++) {
            if (rignthEdge > i)
                r[i] = Math.min(r[2*midIndex - i], rignthEdge - i);
            else
                r[i] = 1;
            while (i - r[i] >= 0 && i + r[i] <= s_temp.length()-1 &&s_temp.charAt(i - r[i]) == s_temp.charAt(i + r[i])) {
                r[i]++;
            }
            if (i + r[i] > rignthEdge) {
                midIndex = i;
                rignthEdge = i+r[i];
            }

            if (2 * r[i] - 1 > maxLen) {
                maxLen = 2*r[i]-1;
                initIndex = i-r[i]+1;
            }
        }
        String res = "";
        for (int i=initIndex; i<initIndex+maxLen; i++) {
            if (s_temp.charAt(i) != '#')
                res += s_temp.charAt(i);
        }
        return res;

    }
}