LeetCode 笔记四 (2019/09/26)

Longest Palindromic Substring

Given a string s, find the longest palindromic substring in s. You may assume that the maximum length of s is 1000.

简单点说就是判断回文。

Example

example1

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

example2

Input: “cbbd”
Output: “bb”

Code

之前几道题基本上都用字典可以很高效的完成，于是一开始我也用了字典：

class Solution:
	def longestPalindrome(self, s: str) -> str:
        if len(s) == 0:
            return s
        Dict = {}
        r = ''
        for i, c in enumerate(s):
            if (c in Dict):
                if (s[Dict[c]:i+1] == s[Dict[c]:i+1][::-1]) and (len(r) < len(s[Dict[c]:i+1])):
                    r = s[Dict[c]:i+1]
            else:
                Dict[c] = i
        if len(r) == 0:
            return s[0]
        else:
            return r

仔细看这个代码，会发现这个代码是错的。这里是将str作为字典的键，在这个代码中，只保存了第一次记录到同一个str的索引，这样无法同时记录中间多个索引，提交果然报错了，看错误的地方比较好理解一点：

Input: "babadada"
Output: "bab"
Expected: "adada"

于是我将索引保存为字典的键，对代码进行了修改：

class Solution:
    def get_key(self, dct, value):
        return list(filter(lambda x:dct[x] == value, dct))
    
    def longestPalindrome(self, s: str) -> str:
        if len(s) == 0:
            return s
        Dict = {}
        r = ''
        for i, c in enumerate(s):
            if c in Dict.values():
                temp = self.get_key(Dict, c)
                for j in temp:
                    if (s[j:i+1] == s[j:i+1][::-1]) and (len(r) < len(s[j:i+1])):
                        r = s[j:i+1]
            Dict[i] = c
        
        if len(r) == 0:
            return s[0]
        else:
            return r

这次提交正确了，不过效果是真的不是很好：

103 / 103 test cases passed.
Runtime: 6544 ms
Memory Usage: 13.8 MB

Runtime: 6544 ms, faster than 13.06% of Python3 online submissions for Longest Palindromic Substring.
Memory Usage: 13.8 MB, less than 22.69% of Python3 online submissions for Longest Palindromic Substring.

其实看这段代码，可以发现这代码用了两次循环…还不如不用字典，直接循环遍历：

class Solution:
    def longestPalindrome(self, s: str) -> str:
        if len(s) == 0:
            return s
        r = ''
        for i in range(len(s)):
            for j in range(len(s), i, -1):
                if len(r) >= j-i:
                    break
                elif s[i:j] == s[i:j][::-1]:
                    r = s[i:j]
                    break
        return r

这样看起来还比较容易理解，提交结果还比字典好了一丢丢：

103 / 103 test cases passed.
Runtime: 5596 ms
Memory Usage: 13.9 MB

Runtime: 5596 ms, faster than 17.25% of Python3 online submissions for Longest Palindromic Substring.
Memory Usage: 13.9 MB, less than 22.69% of Python3 online submissions for Longest Palindromic Substring.

一点题外话…每天都忘记过来更新…继续冲啊！再过一年半就要开始为了找工作而血战，为了生存，用《杀手界》的一句话来说，只有拼啦！