From 01490da048e455b67f8919873f6fc7fb851a03d6 Mon Sep 17 00:00:00 2001
From: Asterisk <44215173+casnz1601@users.noreply.github.com>
Date: Mon, 11 Oct 2021 13:30:42 +0800
Subject: [PATCH] =?UTF-8?q?Update=200093.=E5=A4=8D=E5=8E=9FIP=E5=9C=B0?=
 =?UTF-8?q?=E5=9D=80.md?=
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python3代码修正和补充注释
---
 problems/0093.复原IP地址.md | 65 ++++++++++++++++++++-------------
 1 file changed, 40 insertions(+), 25 deletions(-)

diff --git a/problems/0093.复原IP地址.md b/problems/0093.复原IP地址.md
index 4e56ddc4..1d8ad296 100644
--- a/problems/0093.复原IP地址.md
+++ b/problems/0093.复原IP地址.md
@@ -342,32 +342,47 @@ class Solution:
 ```
 
 python3:
-```python
-class Solution(object):
-    def restoreIpAddresses(self, s):
-        """
-        :type s: str
-        :rtype: List[str]
-        """
-        ans = []
-        path = []
-        def backtrack(path, startIndex):
-            if len(path) == 4:
-                if startIndex == len(s):
-                    ans.append(".".join(path[:]))
-                    return
-            for i in range(startIndex+1, min(startIndex+4, len(s)+1)):  # 剪枝
-                string = s[startIndex:i]
-                if not 0 <= int(string) <= 255:
-                    continue
-                if not string == "0" and not string.lstrip('0') == string:
-                    continue
-                path.append(string)
-                backtrack(path, i)
-                path.pop()
+```python3
+class Solution:
+    def __init__(self):
+        self.result = []
 
-        backtrack([], 0)
-        return ans```
+    def restoreIpAddresses(self, s: str) -> List[str]:
+        '''
+        本质切割问题使用回溯搜索法，本题只能切割三次，所以纵向递归总共四层
+        因为不能重复分割，所以需要start_index来记录下一层递归分割的起始位置
+        添加变量point_num来记录逗号的数量[0,3]
+        '''
+        self.result.clear()
+        if len(s) > 12: return []
+        self.backtracking(s, 0, 0)
+        return self.result
+
+    def backtracking(self, s: str, start_index: int, point_num: int) -> None:
+        # Base Case
+        if point_num == 3:
+            if self.is_valid(s, start_index, len(s)-1):
+                self.result.append(s[:])
+            return
+        # 单层递归逻辑
+        for i in range(start_index, len(s)):
+            # [start_index, i]就是被截取的子串
+            if self.is_valid(s, start_index, i):
+                s = s[:i+1] + '.' + s[i+1:]
+                self.backtracking(s, i+2, point_num+1)  # 在填入.后，下一子串起始后移2位
+                s = s[:i+1] + s[i+2:]    # 回溯
+            else:
+                # 若当前被截取的子串大于255或者大于三位数，直接结束本层循环
+                break
+    
+    def is_valid(self, s: str, start: int, end: int) -> bool:
+        if start > end: return False
+        # 若数字是0开头，不合法
+        if s[start] == '0' and start != end:
+            return False
+        if not 0 <= int(s[start:end+1]) <= 255:
+            return False
+        return True
 ```