LeetCode 5. Longest Palindromic Substring Python3
Description
點選檢視題目
Given a string
s, find the longest palindromic substring in
s. You may assume that the maximum length of s is 1000.
Example 1:
Input: “babad”
Output: “bab”
Note: “aba” is also a valid answer.
Example 2:
Input: “cbbd”
Output: “bb”
問題分析
這道題是一道
中等題,題目的意思是給定一個字元串,找出此字元串的最長回文子串,學習過程式設計的人應該對回文字元串比較了解吧!對于這道題,我最先想到的就是暴力搜尋解法,枚舉出此字元串的所有子串,然後判斷是不是回文字元串,我大緻算了下複雜度,是O(n3),我感覺應該會逾時,果然逾時了。。。。
我在浏覽評論區的時候,發現了一個大佬寫的暴力搜尋代碼,非常簡潔,而且還通過了,不由得感慨自己的代碼寫的太菜了,自己還需要多加努力。
其實這道題要考察的是動态規劃,之前沒有認真學過動态規劃,覺得自己現階段解釋也解釋不太清楚,但是看别人的代碼也能了解咋回事,是以就不在這裡解釋了,通過代碼了解就好。
代碼實作
逾時的暴力搜尋代碼
class Solution:
def longestPalindrome(self, s: str) -> str:
for i in range(len(s) - 1, -1, -1):
j = 0
while j + i < len(s):
temp = s[j:j + i + 1]
j += 1
if self.judge(temp):
return temp
return ""
def judge(self, s: str) -> bool:
for i in range(len(s) // 2):
if s[i] != s[len(s) - i - 1]:
return False
return True
逾時的案例:"aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaabcaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa"
通過的暴力搜尋代碼:
class Solution:
def longestPalindrome(self, s: str) -> str:
if not s:
return ""
for i in range(len(s), 0, -1):
for j in range(len(s) - i + 1):
if s[j:j + i] == s[j:j + i][::-1]:
return s[j:j+i]
動态規劃實作
class Solution:
def longestPalindrome(self, s: str) -> str:
if not s:
return ""
length = len(s)
dp = [[False] * length for _ in range(length)]
left = 0
right = 0
for i in range(length - 2, -1, -1):
dp[i][i] = True
for j in range(i+1, length):
dp[i][j] = (s[i] == s[j]) and (j - i < 3 or dp[i + 1][j - 1])
# j - i < 3 時,其實j - i要麼等于1(aa)要麼等于2(aba),無論哪種情況都滿足
if dp[i][j] and right - left < j - i:
right = j
left = i
return s[left:right + 1]
運作結果對比
從圖中可以看出,自己寫的代碼遠不如Python自帶的高效,是以要多看别人寫的優秀代碼,以提高自己的程式設計能力。
ok!大功告成了,如果你有其他的方法或者問題,歡迎在評論區交流。