# 数字字符串转化成IP地址
 [](https://imhuay.gitbook.io/studies/algorithms/problems/2022/01/pages/R5NyOzkn3qAZy7wCx1pS#中等) [](https://imhuay.gitbook.io/studies/algorithms/problems/2022/01/pages/R5NyOzkn3qAZy7wCx1pS#牛客) [](https://imhuay.gitbook.io/studies/algorithms/problems/2022/01/pages/R5NyOzkn3qAZy7wCx1pS#深度优先搜索)
**问题简述**
```
给定只包含数字的字符串，将该字符串转化成 IP 地址的形式，返回所有可能的情况。
例如：给出的字符串为"25525522135",
返回：["255.255.22.135", "255.255.221.35"]
```
> [数字字符串转化成IP地址\_牛客题霸\_牛客网](https://www.nowcoder.com/practice/ce73540d47374dbe85b3125f57727e1e)
**思路：DFS + 回溯**
* 相当于构建一颗三叉树；
Python
```python
#
# 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
#
#
# @param s string字符串
# @return string字符串一维数组
#
class Solution:
def restoreIpAddresses(self , s: str) -> List[str]:
# write code here
ret = []
def valid(x):
"""验证函数"""
if not x: # 非空
return False
if x.startswith('0') and len(x) > 1: # 存在前缀 0
return False
return int(x) <= 255
def dfs(k, depth, tmp):
if depth == 3: # 到第三层的时候，直接判断所有剩余字符
if valid(s[k:]):
tmp.append(s[k:])
ret.append('.'.join(tmp))
tmp.pop() # 这里也要回溯
return
for i in range(1, 4):
sub = s[k: k + i]
if valid(sub):
tmp.append(sub)
dfs(k + i, depth + 1, tmp)
tmp.pop()
dfs(0, 0, [])
return ret
```
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