[LeetCode] 5. Longest Palindromic Substring _Medium tag: Two pointers
2021-07-29 00:38 Johnson_强生仔仔 阅读(10) 评论(0) 编辑 收藏 举报Given a string s
, return the longest palindromic substring in s
.
Example 1:
Input: s = "babad" Output: "bab" Note: "aba" is also a valid answer.
Example 2:
Input: s = "cbbd" Output: "bb"
Example 3:
Input: s = "a" Output: "a"
Example 4:
Input: s = "ac" Output: "a"
Constraints:
1 <= s.length <= 1000
s
consist of only digits and English letters (lower-case and/or upper-case),
Ideas:
1. 利用 来建一个palin, T: O(n * n) , S: O(n * n), 然后再用两个for loop 去看是否为palin,再更新ans及ans_length, 但是Leetcode上Time Limit Exceeded.
Code
class Solution: def longestPalindrome(self, s: str) -> str: palin = self.generatePalin(s) n = len(s) ans_length, ans = 1, s[0] for i in range(n): for j in range(i, n): if palin[i][j] and j - i + 1 > ans_length: ans_length, ans = j - i + 1, s[i : j + 1] return ans def generatePalin(self, s): n = len(s) palin = [[False] * n for _ in range(n)] for i in range(n): palin[i][i] = True if i > 0: palin[i - 1][i] = s[i] == s[i - 1] for width in range(2, n): for start in range(n): if start + width < n and s[start] == s[start + width]: palin[start][start + width] = palin[start + 1][start + width - 1] return palin
Idea 2: 用一个helper function, 每次往两边衍生, 如果当前是palin的话,然后再更新ans
Code
class Solution: def longestPalindrome(self, s: str) -> str: #palin = self.generatePalin(s) n = len(s) ans =s[0] for i in range(n): # odd case, ex: "aba" odd = self.helper(s, i, i) if len(odd) > len(ans): ans = odd # even case even = self.helper(s, i, i + 1) if len(even) > len(ans): ans = even return ans def helper(self, s, l, r): while l >=0 and r <len(s) and s[l] == s[r]: l-=1 r+=1 return s[l+1:r]