Update 005._longest_palindromic_substring.md

Author: 0xkookoo

docs/Leetcode_Solutions/Python/005._longest_palindromic_substring.md

@@ -45,28 +45,28 @@ LCSuff(S1...p, T1...q) = LCS(S1...p1, T1...q-1) if S[p] = T[q] else 0
 https://en.wikibooks.org/wiki/Algorithm_Implementation/Strings/Longest_common_substring#Python_2
 
 这样也超时？
-```python
-    class Solution(object):
-        def longestPalindrome(self, s):
-            """
-            :type s: str
-            :rtype: str
-            """
-            def lcs(s1, s2):
-                m = [[0] * (1 + len(s2)) for i in xrange(1 + len(s1))]
-                longest, x_longest = 0, 0
-                for x in xrange(1, 1 + len(s1)):
-                    for y in xrange(1, 1 + len(s2)):
-                        if s1[x - 1] == s2[y - 1]:
-                            m[x][y] = m[x - 1][y - 1] + 1
-                            if m[x][y] > longest:
-                                longest = m[x][y]
-                                x_longest = x
-                        else:
-                            m[x][y] = 0
-                return s1[x_longest - longest: x_longest]
-    
-            return lcs(s, s[::-1])
+```
+class Solution(object):
+    def longestPalindrome(self, s):
+        """
+        :type s: str
+        :rtype: str
+        """
+        def lcs(s1, s2):
+            m = [[0] * (1 + len(s2)) for i in xrange(1 + len(s1))]
+            longest, x_longest = 0, 0
+            for x in xrange(1, 1 + len(s1)):
+                for y in xrange(1, 1 + len(s2)):
+                    if s1[x - 1] == s2[y - 1]:
+                        m[x][y] = m[x - 1][y - 1] + 1
+                        if m[x][y] > longest:
+                            longest = m[x][y]
+                            x_longest = x
+                    else:
+                        m[x][y] = 0
+            return s1[x_longest - longest: x_longest]
+
+        return lcs(s, s[::-1])
 ```
 因为以为这样s[::-1]已经很快了.
 
@@ -98,92 +98,90 @@ we check if the substring’s indices are the same as the reversed substring’s
 
 参考https://shenjie1993.gitbooks.io/leetcode-python/content/005%20Longest%20Palindromic%20Substring.html
 ```python
-    class Solution(object):
-        def longestPalindrome(self, s):
-            """
-            :type s: str
-            :rtype: str
-            """
-            n = len(s)
-    
-            # empty or one char
-            if n < 2:
-                return s
-    
-            # left index of the target substring
-            l = 0
-            # right index of the target substring
-            r = 0
-            # length of the longest palindromic substring for now
-            m = 0
-            # length of the current substring
-            c = 0
-    
-            # Whether the substring contains the first character or last character and is palindromic
+class Solution(object):
+    def longestPalindrome(self, s):
+        """
+        :type s: str
+        :rtype: str
+        """
+        n = len(s)
+
+        # empty or one char
+        if n < 2:
+            return s
+
+        # left index of the target substring
+        l = 0
+        # right index of the target substring
+        r = 0
+        # length of the longest palindromic substring for now
+        m = 0
+        # length of the current substring
+        c = 0
+
+        # Whether the substring contains the first character or last character and is palindromic
+        b = True
+        for i in range(n):
+            # Odd situation
+            for j in range(min(n-i,i+1)):
+                if s[i-j] != s [i+j]:
+                    b = False
+                    break
+                else:
+                    c = 2 * j + 1
+
+            if c > m :
+                l = i - j + 1 - b
+                r = i + j + b
+                m = c 
             b = True
-            for i in range(n):
-                # Odd situation
-                for j in range(min(n-i,i+1)):
-                    if s[i-j] != s [i+j]:
-                        b = False
-                        break
-                    else:
-                        c = 2 * j + 1
-    
-                if c > m :
-                    l = i - j + 1 - b
-                    r = i + j + b
-                    m = c 
-                b = True
-    
-                # Even situation
-                for j in range(min(n - i - 1, i + 1)):
-                    if (s[i - j] != s[i + j + 1]):
-                        b = False
-                        break
-                    else:
-                        c = 2 * j + 2
-                if (c > m):
-                    l = i - j + 1 - b
-                    r = i + j + 1 + b
-                    m = c
-                b = True
-            return s[l:r]
+
+            # Even situation
+            for j in range(min(n - i - 1, i + 1)):
+                if (s[i - j] != s[i + j + 1]):
+                    b = False
+                    break
+                else:
+                    c = 2 * j + 2
+            if (c > m):
+                l = i - j + 1 - b
+                r = i + j + 1 + b
+                m = c
+            b = True
+        return s[l:r]
 ```
 以上是参考版本，自己写的版本：
 ```python
-    class Solution(object):
-        def longestPalindrome(self, s):
-            """
-            :type s: str
-            :rtype: str
-            """
-            n = len(s)
-    
-            m,l,r = 0,0,0
-    
-            for i in range(n):
-                # odd case
-                for j in range(min(i+1,n-i)):
-                    if s[i-j] != s[i+j]:
+class Solution(object):
+    def longestPalindrome(self, s):
+        """
+        :type s: str
+        :rtype: str
+        """
+        n = len(s)
+
+        m,l,r = 0,0,0
+
+        for i in range(n):
+            # odd case
+            for j in range(min(i+1,n-i)):
+                if s[i-j] != s[i+j]:
+                    break
+                if 2*j + 1 > m :
+                    m = 2 * j + 1
+                    l = i-j
+                    r = i+j
+
+
+            if i+1 < n and s[i] == s[i+1]:
+                for j in range(min(i+1,n-i-1)):
+                    if s[i-j] != s[i+j+1]:
                         break
-                    if 2*j + 1 > m :
-                        m = 2 * j + 1
+                    if 2 * j + 2 > m :
+                        m = 2*j +2
                         l = i-j
-                        r = i+j
-    
-    
-                if i+1 < n and s[i] == s[i+1]:
-                    for j in range(min(i+1,n-i-1)):
-                        if s[i-j] != s[i+j+1]:
-                            break
-                        if 2 * j + 2 > m :
-                            m = 2*j +2
-                            l = i-j
-                            r = i+j+1
-    
-    
-            return s[l:r+1]
+                        r = i+j+1
+        return s[l:r+1]
 ```