【重点】【DP】5.最长回文子串|516.最长回文子序列

两个求解目标类似的题目，对比记忆！

5.最长回文子串

题目

法1：二维DP

最基础方法！必须掌握！
O(N^2) + O(N^2)

class Solution {
    public String longestPalindrome(String s) {
        int n = s.length();
        if (n == 1) {
            return s;
        }
        String res = s.substring(0, 1);
        boolean[][] dp = new boolean[n][n];
        for (int i = 0; i < n; ++i) {
            dp[i][i] = true;
        }
        for (int i = n - 1; i >= 0; --i) {
            for (int j = i + 1; j < n; ++j) {
                dp[i][j] = (j - i == 1 || dp[i + 1][j - 1]) && (s.charAt(i) == s.charAt(j));
                if (dp[i][j] && (j - i + 1 > res.length())) {
                    res = s.substring(i, j + 1);
                } 
            }
        }

        return res;
    }
}

法2：中心扩展法

O(N^2) + O(1)

class Solution {
    public String longestPalindrome(String s) {
        if (s.length() < 2) {
            return s;
        }
        String res = "";
        for (int i = 0; i < s.length(); ++i) {
            String res1 = palindrome(s, i, i);
            String res2 = palindrome(s, i, i + 1);
            res = res1.length() > res.length() ? res1 : res;
            res = res2.length() > res.length() ? res2 : res;
        }

        return res;
    }

    public String palindrome(String s, int i, int j) {
        while (i >= 0 && j < s.length() && (s.charAt(i) == s.charAt(j))) {
            --i;
            ++j;
        }

        return s.substring(i + 1, j);
    }
}

516.最长回文子序列

题目

class Solution {
    public int longestPalindromeSubseq(String s) {
        int n = s.length();
        int[][] dp = new int[n][n];
        for (int i = 0; i < n; ++i) {
            dp[i][i] = 1;
        }
        for (int i = n - 1; i >= 0; --i) {
            for (int j = i + 1; j < n; ++j) {
                if (s.charAt(i) == s.charAt(j)) {
                    dp[i][j] = (j - i == 1 ? 0 : dp[i + 1][j - 1]) + 2;
                } else {
                    dp[i][j] = Math.max(dp[i + 1][j], dp[i][j - 1]);
                }
            }
        }

        return dp[0][n - 1];
    }
}