[LeetCode][python3]0005. Longest Palindromic Substring

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N2I -2020.03.15

1. My first try

class Solution:
    def longestPalindrome(self, s: str) -> str:
        if not s:
            return ""
        elif len(s)==1:
            return s
        l=1
        r=-1
        max_sub=0
        for i in range(1,len(s)):
            cur_l,cur_r=self.lgcheck(s,i)
            if r-l-1<cur_r-cur_l-1:
                l=cur_l
                r=cur_r            
        return s[l+1:r]
            
    
    def lgcheck(self,s,start):
        start_l=start-1
        start_r=start+1
        while start_l>=0 and start_r<len(s):
            if s[start_l]!=s[start_r]:
                break
            else:
                start_l-=1
                start_r+=1
        cur_max=start_r-start_l-1
        r=start_r
        l=start_l
        
        start_r=start
        start_l=start-1
        while  start_l>=0 and start_r<len(s):
            if s[start_l]!=s[start_r]:
                break
            else:
                start_l-=1
                start_r+=1
        if cur_max<start_r-start_l-1:
            r=start_r
            l=start_l

Explanation:

Just a normal solution for beginner.

2. A better solution

class Solution:
    def longestPalindrome(self, s: str) -> str:
        str_len = len(s)
        # if length of string < 2 or s is palindrome already
        if str_len < 2 or s == s[::-1]:
            return s
start, max_len = 0, 1
for i in range(1, str_len):
            odd_start = i - max_len - 1
            even_start = i - max_len
            odd = s[odd_start:i + 1]  # len(odd) = max_len + 2
            even = s[even_start:i + 1]  # len(even) = max_len + 1
if odd_start >= 0 and odd == odd[::-1]:
                start = odd_start
                max_len += 2
            elif even_start >= 0 and even == even[::-1]:
                start = even_start
                max_len += 1
        return s[start:start + max_len]

Note:

• even[::-1] is the reverse array for even[:].
• s[a::b] is a sub_array of s, a stands for the start index (default 0 or -1 when b<0), b stands for what length for a jump.

print("note for s[a::b]")
print("while a stants and b>0")
s=[0,1,2,3,4,5]
s[::1]
>>>[0,1,2,3,4,5]
s[1::]
>>>[1,2,3,4,5]
s[1::2]
>>>[1,3,5]
s[1::3]
>>>[1,4]
s[2::2]
>>>[2,4]
print("while a <len(s) and b<0")
s[::-1]
>>>[5,4,3,2,1,0]
s[::-2]
>>>[5,3,1]
s[-2::-2]
>>>[4,2,0]

Explanation:

This solution use even==even[::-1] to check the selecting area even=s[even_start,i+1]is palindrome or not. Since even or odd will always be 1 or 2 size longer then the current longest substring. start stands for the start of the current longest substring, will only changes when it finds a new longest one in odd or even.