LeetCode Medium：5. Longest Palindromic Substring

一、题目

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

Example:

Input: "babad"

Output: "bab"

Note: "aba" is also a valid answer.

Example:

Input: "cbbd"

Output: "bb"
题目的意思是找最大的回文串，回文串有两种：一种是中间无重复的，如“aba”，一种是中间是重复的。比如“xaaaaax”
二、思路
我能想到的是暴力解法，将每一个字符都作为回文字符串中心，依次扩展，但是这种算法复杂度很高，如代码longestPalindrome0，而且我没有考虑到中间带重复的，所以没有编译通过，在网上看大神的博客，
找了两种比较好的做法，代码longestPalindrome1和代码longestPalindrome2，将中间重复的作为一个整体考虑。
三、代码

#coding:utf-8
def longestPalindrome0(s):
    """
    :type s: str
    :rtype: str
    """
'''
maxcount = 0
    for i in range(1,len(s)-1):
        count = 0
        j = 1
        curleft = i - j
        curright = i + j
        while s[curleft] == s[curright]:
            j+=1
            count+=1
            curleft-=1
            curright+=1
            if curleft < 0 or curright > len(s) - 1:
                print('break')
                break


        if maxcount < count:
            maxcount = count
            a = curleft + 1
            b = curright
            res = s[a:b:1]
    print(maxcount)
    print(res)
    return maxcount,res
'''

'https://www.cnblogs.com/chruny/p/4791078.html'
def longestPalindrome1(s):
    """
    :type s: str
    :rtype: str
    """
    size = len(s)
    if size == 1:
        print(s)
        return s
    if size == 2:
        if s[0] == s[1]:
            print(s)
            return s
        print(s[0])
        return s[0]
    maxp = 1
    ans = s[0]
    i = 0
    while i < size:
        j = i + 1
        while j < size:
            if s[i] == s[j]:
                j += 1
            else:
                break
        k = 0
        while i - k - 1 >= 0 and j + k <= size - 1:
            if s[i - k - 1] != s[j + k]:
                break
            k += 1
        if j - i + 2 * k > maxp:
            maxp = j - i + 2 * k
            ans = s[i - k:j + k]
        if j + k == size - 1:
            break
        i = j
    print(ans)
    return ans

'https://blog.csdn.net/shiroh_ms08/article/details/26094141'
def longestPalindrome3(s):
    if not s or len(s) == 1:
        return s
    # record the result value
    max_length = 1
    start_index = 0
    end_index = 0
    for i in range(0, len(s)-1):
        count = 1
        # aba
        if s[i] != s[i+1]:
            while i-count >= 0 and i + count < len(s) and s[i-count] == s[i+count]:
                count += 1
            if (count-1) * 2 + 1 > max_length:
                max_length = (count-1) * 2 + 1
                start_index = i - count + 1
                end_index = i + count - 1
        # xaaaaax
        else:
            count_repeat = 1
            count = 0
            while i+1 < len(s) and s[i] == s[i+1]:
                i += 1
                count_repeat += 1
            while i-count_repeat+1-count >= 0 and i + count < len(s) and s[i-count_repeat+1-count] == s[i+count]:
                count += 1
            if (count-1) * 2 + count_repeat > max_length:
                max_length = (count-1) * 2 + count_repeat
                start_index = i - count -count_repeat + 2
                end_index = i + count - 1
    print(s[start_index:end_index+1])
    return s[start_index:end_index+1]

if __name__ == '__main__':
    s1 = 'dabbac'
    #longestPalindrome1(s1)
    s3 ='axaaaaax'
    longestPalindrome3(s3)

　参考博客：

https://blog.csdn.net/shiroh_ms08/article/details/26094141

https://www.cnblogs.com/chruny/p/4791078.html