【重来】【vip】5. Longest Palindromic Substring【m】【30】【97】

Given a string S, find the longest palindromic substring in S. You may assume that the maximum length of S is 1000, and there exists one unique longest palindromic substring.

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注释掉的是n^2的

没注释的是线性的算法

这是discuss里面给的解释

还得再看一遍。。

Basic thought is simple. when you increase s by 1 character, you could only increase maxPalindromeLen by 1 or 2, and that new maxPalindrome includes this new character. Proof: if on adding 1 character, maxPalindromeLen increased by 3 or more, say the new maxPalindromeLen is Q, and the old maxPalindromeLen is P, and Q>=P+3. Then it would mean, even without this new character, there would be a palindromic substring ending in the last character, whose length is at least Q-2. Since Q-2 would be >P, this contradicts the condition that P is the maxPalindromeLen without the additional character.

So, it becomes simple, you only need to scan from beginning to the end, adding one character at a time, keeping track of maxPalindromeLen, and for each added character, you check if the substrings ending with this new character, with length P+1 or P+2, are palindromes, and update accordingly.

Now, this is O(n^2) as taking substrings and checking palindromicity seem O(n) time. We can speed up it by realizing that strings are immutable, and there are memory slicing tricks will help to speed these operations up. comparing string equality with "==" is O(1), and using slicing to substring and reverse is ̶a̶l̶s̶o̶ ̶O̶(̶1̶)̶ ̶(̶n̶o̶t̶ ̶t̶o̶t̶a̶l̶l̶y̶ ̶s̶u̶r̶e̶ ̶a̶b̶o̶u̶t̶ ̶t̶h̶e̶ ̶s̶l̶i̶c̶i̶n̶g̶ ̶t̶h̶o̶u̶g̶h̶.̶ ̶ ̶I̶ ̶t̶h̶i̶n̶k̶ ̶i̶t̶ ̶i̶s̶ ̶O̶(̶1̶)̶,̶ ̶b̶u̶t̶ ̶c̶o̶u̶l̶d̶ ̶n̶o̶t̶ ̶f̶i̶n̶d̶ ̶a̶n̶y̶ ̶s̶o̶l̶i̶d̶ ̶l̶i̶t̶e̶r̶a̶t̶u̶r̶e̶ ̶a̶b̶o̶u̶t̶ ̶i̶t̶.̶ O(n) (thanks to ChuntaoLu). But as slicing is optimized by the interpreter's C code, it should run pretty fast. I'm pretty new to Python. Would appreciate you would give more insights or further optimization.

Thus, here is the O(n) method:

class Solution(object):
    def longestPalindrome(self, s):
        
        if len(s)==0:
            return 0
        maxLen=1
        start=0
        for i in xrange(len(s)):
            print i,s[i],s[start:start+maxLen]
            if i-maxLen >=1 and s[i-maxLen-1:i+1]==s[i-maxLen-1:i+1][::-1]:
                print s[i-maxLen-1:i+1]
                start=i-maxLen-1
                maxLen+=2
                continue

            if i-maxLen >=0 and s[i-maxLen:i+1]==s[i-maxLen:i+1][::-1]:
                start=i-maxLen
                maxLen+=1
        return s[start:start+maxLen]
        
        
        
        '''这是n^2的解法
        def helper(s,l,r):
            while l>= 0 and r< len(s) and s[l] == s[r]:
                l-=1
                r+=1
            return s[l+1:r]
            
            
        maxx = 0
        res = ''
        
        for i in xrange(len(s)):
            tmp = helper(s,i,i)
            if len(tmp) > len(res):
                res = tmp
            tmp = helper(s,i,i+1)
            
            if len(tmp) > len(res):
                res = tmp
        return res
        '''
        
        '''
        for i in xrange(len(s)):
            for j in xrange(len(s)-i -1,maxx,-1):
                #print i,j,s[i:i+j]
                
                #print i,j
                
                if s[i] != s[i+j]:
                    continue
                pos = 0
                while pos <= j/2 and s[i+pos] == s[i+j-pos]:
                    pos += 1
                if pos == j/2 + 1 and maxx < j:
                    res = s[i:i+j+1]
                    maxx = j
                    
                
                t = s[i:i+j]
                tt = t[::-1]
                
                if t == tt:
                    if j > maxx:
                        res = t
                        maxx = max(maxx,j)
                
        return res
        '''
        
        """
        :type s: str
        :rtype: str
        """