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chore(docs): sync doc metadata [skip ci]
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app/docs/CommunityShare/Leetcode/93复原Ip地址.md

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---
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title: 93. Restore IP Addresses
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date: '2025/3/25-14:03'
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date: "2025/3/25-14:03"
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tags:
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- - Python
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- - Answer
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abbrlink: 9d0d3b9c
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docId: d5evrnoglwjvmyginjq84bl0
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---
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# QUESTION:
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One important note: `parts + [part]` is pass-by-value, not by reference like in LeetCode 46.
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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.
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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.
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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.
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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.
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`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
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# Code:
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```python
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backtrack(0, [])
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return res
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```
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```

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