5. Longest Palindromic Substring

问题

求一个字符串的最长回文子串

Input: "babad"
Output: "bab"
Note: "aba" is also a valid answer.

思路

这道题跟647. Palindromic Substrings差不多，一个是求子回文串数量，一个是求最长子回文串。

只要把Palindromic Substrings的代码改写一下，把求数量改成求最长就可以了。

主要用了两种方法，一个是朴素的做法，按每个字符从两边扩张，另一个是著名的马拉车算法。具体思路在Palindromic Substrings中已经分析过了，此处不多阐述。

代码

朴素做法：时间复杂度O(n^2)，空间复杂度O(1)

class Solution(object):
    def longestPalindrome(self, s):
        """
        :type s: str
        :rtype: int
        """
        n = len(s)
        maxI = 0
        maxLen = 0
        for i in range(2*n-1):
            left = i/2
            right = left + i%2
            while(left >= 0 and right < n and s[left]==s[right] ):
                left -= 1
                right += 1
            if(right - left -1 > maxLen):
                maxI = left + 1
                maxLen = right - left -1
        return s[maxI:maxI+maxLen]

马拉车算法：时间复杂度O(n)，空间复杂度O(1)

class Solution(object):
    def longestPalindrome(self, s):
        """
        :type s: str
        :rtype: int
        """
        s = '#' + '#'.join(s) + '#'
        n = len(s)
        p = [0]*n
        id = mx = 0
        maxR = 0
        maxI = 0
        for i in range(1,n-1):
            if(mx > i):
                p[i] = min(mx-i, p[2*id-i])
            while i+p[i]+1 < n and i-p[i]-1 >=0 and s[i+p[i]+1] == s[i-p[i]-1]:
                p[i] += 1
            if(i + p[i] > mx):
                id = i
                mx = i + p[i]
            if(p[i]>maxR ):
                maxR = p[i]
                maxI = i
        return s[maxI - maxR + 1: maxI + maxR].replace('#','')

类似题目

647. Palindromic Substrings

516. Longest Palindromic Subsequence