LeetCode - Longest Palindromic Substring

LeetCode - Longest Palindromic Substring

Ths problem is described as following :

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.

方法一：

起初，我采用较为暴力的方法，查找每个可能的回文字串,时间复杂度 O(N3) ,空间复杂度 O(1) 。
具体而言，从每个元素i开始查找，固定开始元素i,从最后一个元素开始找相同字符j，若i~j长度大于当前最长字串的长度，就比较其中间元素，如果是回文的，就更新最长字串，并记录长度。如此往复，当开始元素下标i到最后一个元素的长度都小于当前最长字串的长度，就结束算法。这是一种暴力的方法，但是注意做到了一些优化，然而在测试时，遇到极端情况（1000个a中间插入bc）还是较为消耗时间，所以放弃这一方法，然而代码如下：

class Solution:
    # @return a string
    def longestPalindrome(self, s):
        longest = 0  # to store the length of the longest Palindrome
        longstr = ''
        i = 0
        while i < len(s) and len(s)-i > longest:
            j = len(s)-1  # from the last character
            while j > i:
                if s[j] == s[i] and longest < j-i+1:
                    l = i
                    r = j
                    while l < r and s[l] == s[r]:
                        l += 1
                        r -= 1
                    if not l < r:
                        longest = j-i+1
                        longstr = s[i:j+1]
                        break
                j -= 1
            i += 1
        return longstr

方法二：

之后考虑中心扩展的方法：时间复杂度 O(N2) ,空间复杂度 O(1)
这一方法，要注意中心是奇数和偶数两种情况分别讨论。具体编码时，中心值从整个string的中间元素往两侧扩展效率较高，代码如下.

class Solution:
    # @return a string
    def longestPalindrome(self, s):
        if len(s) == 0:
            return ''
        longest, longstr = 0, ''
        i, j = (len(s)-1)/2, len(s)/2
        while i >= 0 and j < len(s):
            args = [(s,i,i), (s, i, i+1), (s, j, j), (s,j,j+1)]
            for arg in args:
                stemp = self.PalindromewithCenter(*arg)
                if len(stemp) > longest:
                    longest, longstr = len(stemp), stemp
            if longest >= i*2:
                break
            i, j = i-1, j+1
        return longstr

    def PalindromewithCenter(self, s, l, r):
        while l >= 0 and r <= len(s)-1 and s[l] == s[r]:
            l -= 1
            r += 1
        return s[l+1:r]