Given a string s, find the longest palindromic substring in s. You may assume that the maximum length of s is 1000.
Example:
Input: "babad" Output: "bab" Note: "aba" is also a valid answer.
Example:
Input: "cbbd" Output: "bb"
manacher 算法
回文有奇数偶数的问题,所以加上gap,这样字符串一定是奇数,所以只考虑奇数匹配就行。
a b c -----> #a#b#c#
a b -->#a#b#
1、预处理成上面的样子,为处理方便,在最前面加一个从未出现的字符$
2、建立数组P,P[i] 来记录字符S[i]为中心的最长回文子串向左/向右扩张的长度(包括S[i])
例如:
12212321
预处理:$#1#2#2#1#2#3#2#1#
P: 12125214121612121
绿框之外的暴力
1 class Solution: 2 def longestPalindrome(self, s): 3 """ 4 :type s: str 5 :rtype: str 6 """ 7 # preprocess 8 slist = list(s) 9 for i in range((len(s) + 1) * 2)[::2]: 10 slist.insert(i, "#") 11 slist.insert(0, '$') 12 13 p = self.manacher(slist) 14 15 i = p.index(max(p)) 16 ans = ''.join(slist[i-p[i]+1:i+p[i]]) 17 return ans.replace('#','').replace('$','') 18 def manacher(self, slist): 19 20 21 # 计算p 22 p = [0] * len(slist) 23 p[0] = 1 24 id = 0 25 mx = 1 26 print(slist) 27 for i in range(1,len(slist)): 28 if mx > i: 29 p[i] = min(p[id * 2 - i], mx - i) 30 else: 31 p[i] = 1 32 33 #暴力 34 while i +p[i]<len(slist) and slist[i - p[i]]==slist[i + p[i]]: 35 p[i] = p[i]+1 36 #更新最大三元组 37 if(mx < i + p[i]): 38 mx = i + p[i] 39 id = i 40 41 return p
动态规划
此题还可以用动态规划Dynamic Programming来解,我们维护一个二维数组dp,其中dp[i][j]表示字符串区间[i, j]是否为回文串,当i = j时,只有一个字符,肯定是回文串,如果i = j + 1,说明是相邻字符,此时需要判断s[i]是否等于s[j],如果i和j不相邻,即i - j >= 2时,除了判断s[i]和s[j]相等之外,dp[j + 1][i - 1]若为真,就是回文串,通过以上分析,可以写出递推式如下:
dp[i, j] = 1 if i == j
= s[i] == s[j] if j = i + 1
= s[i] == s[j] && dp[i + 1][j - 1] if j > i + 1
中心扩散法: