chore: add daily leetcode post 93复原Ip地址

Author: longsizhuo

app/docs/CommunityShare/Leetcode/93复原Ip地址.md

@@ -0,0 +1,67 @@
+---
+title: 93. Restore IP Addresses
+date: '2025/3/25-14:03'
+tags:
+  - - Python
+  - - Answer
+abbrlink: 9d0d3b9c
+---
+
+# QUESTION：
+
+[93. Restore IP Addresses](https://leetcode.cn/problems/restore-ip-addresses/description/?envType=company&envId=mihoyo&favoriteSlug=mihoyo-all)
+
+# My Think：
+
+This MiHoYo coding test question is very similar to LeetCode 46 (Permutations), and both rely on the backtracking approach.
+
+As shown in the code, during the first traversal, we may get something like `['2', '5', '5', '2']` as our initial parts, but at this point, we haven’t traversed the entire string yet.
+
+When we enter the next level of DFS, the pointer moves forward by +length, so it effectively moves to the far right of the current segment.
+If the pointer has reached the end of the string, it means we’ve visited all characters — in this case, we’ve found one valid answer and can add it to the result list.
+
+One important note: `parts + [part]` is pass-by-value, not by reference like in LeetCode 46.
+This means we don’t need to manually undo changes (i.e., no need to backtrack with pop()), because each recursive call creates a new list.
+
+
+现在在做的是MiHoyo的笔试题, 这个和46.permutation非常相似, 都是回溯思想.
+
+如代码所示, 在第一遍遍历中, 我们会得到`['2','5','5','2']` 作为第一个parts, 但是我们并没有遍历完整个字符串.
+
+这个时候如果我们进入新的dfs的时候, 我们的指针因为刚刚`+length`, 所以我们实际上
+指到的是最右边,所以当当前指针已经指到最右边的时候, 说明我们遍历完了所有字符, 那么这就是答案之一.
+
+`parts + [part]` 是值传递而不是46的引用传递, 所以不用手动撤销更改
+# Code：
+
+```python
+from typing import List
+
+class Solution:
+    def restoreIpAddresses(self, s: str) -> List[str]:
+        res = []
+
+        def backtrack(start: int, parts: List[str]):
+            # 终止条件：正好4段且用完所有字符
+            # Stop condition: exactly 4 segments and all characters used up
+            if len(parts) == 4:
+                if start == len(s):
+                    res.append(".".join(parts))
+                return
+
+            for length in range(1, 4):  # 每段长度1~3 Each segment length 1~3
+                if start + length > len(s):
+                    break
+                part = s[start:start+length]
+
+                # 前导0非法，但0本身合法
+                # Leading 0 is illegal, but 0 itself is legal
+                if len(part) > 1 and part[0] == '0':
+                    continue
+
+                if int(part) <= 255:
+                    backtrack(start + length, parts + [part])  # 注意用 + 避免污染 We need to use + to avoid pollution
+
+        backtrack(0, [])
+        return res
+```