leetcode 最长公共子串

Leetcode 最长公共子串问题可以使用动态规划解决。具体方法是定义一个二维数组，表示两个字符串中以 i,j 为结尾的最长公共子串长度，状态转移方程为：如果 s1[i] = s2[j]，则 dp[i][j] = dp[i-1][j-1] + 1；否则，dp[i][j] = 0。最终，求得的二维数组中的最大值即为最长公共子串的长度。

相关问题

leetcode最长回文子串

根据提供的引用内容，有三种方法可以解决LeetCode上的最长回文子串问题。

方法一是使用扩展中心法优化，即从左向右遍历字符串，找到连续相同字符组成的子串作为扩展中心，然后从该中心向左右扩展，找到最长的回文子串。这个方法的时间复杂度为O(n²)。\[1\]

方法二是直接循环字符串，判断子串是否是回文子串，然后得到最长回文子串。这个方法的时间复杂度为O(n³)，效率较低。\[2\]

方法三是双层for循环遍历所有子串可能，然后再对比是否反向和正向是一样的。这个方法的时间复杂度也为O(n³)，效率较低。\[3\]

综上所述，方法一是解决LeetCode最长回文子串问题的最优解法。
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5. 最长回文子串python

回答: 最长回文子串可以通过两种方法来实现。第一种是使用中心扩展法，代码如下：

class Solution:
    def check(self, s, l, r):
        while l >= 0 and r < len(s) and s[l == s[r]:
            l -= 1
            r += 1
        return l + 1, r - 1

    def longestPalindrome(self, s: str) -> str:
        start, end = 0, 0
        for x in range(len(s)):
            l1, r1 = self.check(s, x, x)
            l2, r2 = self.check(s, x, x + 1)
            if r1 - l1 > end - start:
                start, end = l1, r1
            if r2 - l2 > end - start:
                start, end = l2, r2
        return s[start:end+1]

第二种方法是使用动态规划，代码如下：

class Solution:
    def longestPalindrome(self, s: str) -> str:
        res = ''
        for i in range(len(s)):
            start = max(0, i - len(res) - 1)
            temp = s[start:i+1]
            if temp == temp[::-1]:
                res = temp
            else:
                temp = temp<span class="em">1</span><span class="em">2</span><span class="em">3</span>
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