leetcode-5-最长回文子串*马拉车

方法一：动态规划 O(n2) O(n2)

class Solution:
    def longestPalindrome(self, s: str) -> str:
        size = len(s)
        if size <= 1:
            return s
        # 二维 dp 问题 # 状态：dp[l,r]: s[l:r] 包括 l，r ，表示的字符串是不是回文串
        # 设置为 None 是为了方便调试，看清楚代码执行流程
        dp = [[False for _ in range(size)] for _ in range(size)]
        longest_l = 1
        res = s[0]
        # 因为只有 1 个字符的情况在最开始做了判断
        # 左边界一定要比右边界小，因此右边界从 1 开始
        for r in range(1, size):
            for l in range(r):
                # 状态转移方程：如果头尾字符相等并且中间也是回文
                # 在头尾字符相等的前提下，如果收缩以后不构成区间（最多只有 1 个元素），直接返回 True 即可
                # 否则要继续看收缩以后的区间的回文性
                # 重点理解 or 的短路性质在这里的作用
                if s[l] == s[r] and (r - l <= 2 or dp[l + 1][r - 1]):
                    dp[l][r] = True
                    cur_len = r - l + 1
                    if cur_len > longest_l:
                        longest_l = cur_len
                        res = s[l:r + 1]
                        # 调试语句
                        # for item in dp:
                        #     print(item) # print('---')
        return res

java:

class Solution {
    public String longestPalindrome(String s) {
        if (s == null || s.length() < 2) {
            return s;
        }
        int strLen = s.length();
        int maxStart = 0;  //最长回文串的起点
        int maxEnd = 0;    //最长回文串的终点
        int maxLen = 1;  //最长回文串的长度

        boolean[][] dp = new boolean[strLen][strLen];

        for (int r = 1; r < strLen; r++) {
            for (int l = 0; l < r; l++) {
                if (s.charAt(l) == s.charAt(r) && (r - l <= 2 || dp[l + 1][r - 1])) {
                    dp[l][r] = true;
                    if (r - l + 1 > maxLen) {
                        maxLen = r - l + 1;
                        maxStart = l;
                        maxEnd = r;

                    }
                }

            }

        }
        return s.substring(maxStart, maxEnd + 1);

    }

}

方法二：中心扩展法 O(n^2)

class Solution:
    def longestPalindrome(self, s):
        size = len(s)
        if size == 0:
            return ''
        # 至少是 1
        longest_palindrome = 1
        longest_palindrome_str = s[0]
        for i in range(size):
            palindrome_odd, odd_len = self.__center_spread(s, size, i, i)
            palindrome_even, even_len = self.__center_spread(s, size, i, i + 1)
            # 当前找到的最长回文子串
            cur_max_sub = palindrome_odd if odd_len >= even_len else palindrome_even
            if len(cur_max_sub) > longest_palindrome:
                longest_palindrome = len(cur_max_sub)
                longest_palindrome_str = cur_max_sub
        return longest_palindrome_str
    def __center_spread(self, s, size, left, right):
        """
        left = right 的时候，此时回文中心是一条线，回文串的长度是奇数
        right = left + 1 的时候，此时回文中心是任意一个字符，回文串的长度是偶数
        """
        l = left
        r = right
        while l >= 0 and r < size and s[l] == s[r]:
            l -= 1
            r += 1
        return s[l + 1:r], r - l - 1

java:

class Solution {
    public String longestPalindrome(String s) {
        if (s == null || s.length() < 1) return "";
        int start = 0,end = 0;
        for (int i = 0; i < s.length(); i++){
            int len1 = expandAroundCenter(s,i,i);
            int len2 = expandAroundCenter(s,i,i + 1);
            int len = Math.max(len1,len2);
            if (len > end - start){
                start = i - (len - 1) / 2;
                end = i + len / 2;
            }
        }
        return s.substring(start,end + 1);
    }

    private int expandAroundCenter(String s,int left,int right){
        int L = left, R = right;
        while (L >= 0 && R < s.length() && s.charAt(L) == s.charAt(R)){
            L--;
            R++;
        }
        return R - L -1;
    }
}