From 45863868708ba1c3260060fb5afa854fa6874184 Mon Sep 17 00:00:00 2001
From: jianghongcheng <35664721+jianghongcheng@users.noreply.github.com>
Date: Fri, 26 May 2023 21:03:46 -0500
Subject: [PATCH] =?UTF-8?q?Update=200131.=E5=88=86=E5=89=B2=E5=9B=9E?=
 =?UTF-8?q?=E6=96=87=E4=B8=B2.md?=
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---
 problems/0131.分割回文串.md | 143 ++++++++++++++++++++-----------
 1 file changed, 91 insertions(+), 52 deletions(-)

diff --git a/problems/0131.分割回文串.md b/problems/0131.分割回文串.md
index 30bba455..53f43eea 100644
--- a/problems/0131.分割回文串.md
+++ b/problems/0131.分割回文串.md
@@ -349,12 +349,9 @@ class Solution {
 ```
 
 ## Python 
-**回溯+正反序判断回文串**	
+回溯 基本版
 ```python
 class Solution:
-    def __init__(self):
-        self.paths = []
-        self.path = []
 
     def partition(self, s: str) -> List[List[str]]:
         '''
@@ -363,52 +360,14 @@ class Solution:
         当切割线迭代至字符串末尾，说明找到一种方法
         类似组合问题，为了不重复切割同一位置，需要start_index来做标记下一轮递归的起始位置(切割线)
         '''
-        self.path.clear()
-        self.paths.clear()
-        self.backtracking(s, 0)
-        return self.paths
+        result = []
+        self.backtracking(s, 0, [], result)
+        return result
 
-    def backtracking(self, s: str, start_index: int) -> None:
+    def backtracking(self, s, start_index, path, result ):
         # Base Case
-        if start_index >= len(s):
-            self.paths.append(self.path[:])
-            return
-        
-        # 单层递归逻辑
-        for i in range(start_index, len(s)):
-            # 此次比其他组合题目多了一步判断：
-            # 判断被截取的这一段子串([start_index, i])是否为回文串
-            temp = s[start_index:i+1]
-            if temp == temp[::-1]:  # 若反序和正序相同，意味着这是回文串
-                self.path.append(temp)
-                self.backtracking(s, i+1)   # 递归纵向遍历：从下一处进行切割，判断其余是否仍为回文串
-                self.path.pop()
-            else:
-                continue    
-```
-**回溯+函数判断回文串**
-```python
-class Solution:
-    def __init__(self):
-        self.paths = []
-        self.path = []
-
-    def partition(self, s: str) -> List[List[str]]:
-        '''
-        递归用于纵向遍历
-        for循环用于横向遍历
-        当切割线迭代至字符串末尾，说明找到一种方法
-        类似组合问题，为了不重复切割同一位置，需要start_index来做标记下一轮递归的起始位置(切割线)
-        '''
-        self.path.clear()
-        self.paths.clear()
-        self.backtracking(s, 0)
-        return self.paths
-
-    def backtracking(self, s: str, start_index: int) -> None:
-        # Base Case
-        if start_index >= len(s):
-            self.paths.append(self.path[:])
+        if start_index == len(s):
+            result.append(path[:])
             return
         
         # 单层递归逻辑
@@ -416,9 +375,9 @@ class Solution:
             # 此次比其他组合题目多了一步判断：
             # 判断被截取的这一段子串([start_index, i])是否为回文串
             if self.is_palindrome(s, start_index, i):
-                self.path.append(s[start_index:i+1])
-                self.backtracking(s, i+1)   # 递归纵向遍历：从下一处进行切割，判断其余是否仍为回文串
-                self.path.pop()             # 回溯
+                path.append(s[start_index:i+1])
+                self.backtracking(s, i+1, path, result)   # 递归纵向遍历：从下一处进行切割，判断其余是否仍为回文串
+                path.pop()             # 回溯
             else:
                 continue    
 
@@ -430,9 +389,89 @@ class Solution:
                 return False
             i += 1
             j -= 1
-        return True
+        return True 
 ```
+回溯+优化判定回文函数
+```python
+class Solution:
 
+    def partition(self, s: str) -> List[List[str]]:
+        result = []
+        self.backtracking(s, 0, [], result)
+        return result
+
+    def backtracking(self, s, start_index, path, result ):
+        # Base Case
+        if start_index == len(s):
+            result.append(path[:])
+            return
+        
+        # 单层递归逻辑
+        for i in range(start_index, len(s)):
+            # 若反序和正序相同，意味着这是回文串
+            if s[start_index: i + 1] == s[start_index: i + 1][::-1]:
+                path.append(s[start_index:i+1])
+                self.backtracking(s, i+1, path, result)   # 递归纵向遍历：从下一处进行切割，判断其余是否仍为回文串
+                path.pop()             # 回溯
+            else:
+                continue    
+```
+回溯+高效判断回文子串
+```python
+class Solution:
+    def partition(self, s: str) -> List[List[str]]:
+        result = []
+        isPalindrome = [[False] * len(s) for _ in range(len(s))]  # 初始化isPalindrome矩阵
+        self.computePalindrome(s, isPalindrome)
+        self.backtracking(s, 0, [], result, isPalindrome)
+        return result
+
+    def backtracking(self, s, startIndex, path, result, isPalindrome):
+        if startIndex >= len(s):
+            result.append(path[:])
+            return
+
+        for i in range(startIndex, len(s)):
+            if isPalindrome[startIndex][i]:   # 是回文子串
+                substring = s[startIndex:i + 1]
+                path.append(substring)
+                self.backtracking(s, i + 1, path, result, isPalindrome)  # 寻找i+1为起始位置的子串
+                path.pop()           # 回溯过程，弹出本次已经填在的子串
+
+    def computePalindrome(self, s, isPalindrome):
+        for i in range(len(s) - 1, -1, -1):  # 需要倒序计算，保证在i行时，i+1行已经计算好了
+            for j in range(i, len(s)):
+                if j == i:
+                    isPalindrome[i][j] = True
+                elif j - i == 1:
+                    isPalindrome[i][j] = (s[i] == s[j])
+                else:
+                    isPalindrome[i][j] = (s[i] == s[j] and isPalindrome[i+1][j-1])
+```
+回溯+使用all函数判断回文子串
+```python
+class Solution:
+    def partition(self, s: str) -> List[List[str]]:
+        result = []
+        self.partition_helper(s, 0, [], result)
+        return result
+
+    def partition_helper(self, s, start_index, path, result):
+        if start_index == len(s):
+            result.append(path[:])
+            return
+
+        for i in range(start_index + 1, len(s) + 1):
+            sub = s[start_index:i]
+            if self.isPalindrome(sub):
+                path.append(sub)
+                self.partition_helper(s, i, path, result)
+                path.pop()
+
+    def isPalindrome(self, s):
+        return all(s[i] == s[len(s) - 1 - i] for i in range(len(s) // 2))
+
+```
 ## Go 
 ```go
 var (