# 440. 字典序的第K小数字
## 解法一：dfs暴力（超时）
先dfs回溯法求出字典序数组，然后返回k-1号元素：
```python
class Solution:
def findKthNumber(self, n: int, k: int) -> int:
self.nums = []
for i in range(1, 10): #第一位不能为0
self.dfs(i, n) #从第二位开始dfs回溯
return self.nums[k-1]
def dfs(self, cur, n):
if cur > n: #超出则剪枝
return
self.nums.append(cur)
cur = cur * 10
for i in range(10):
self.dfs(cur + i, n)
```
## 解法二：
参考
```python
class Solution:
def findKthNumber(self, n: int, k: int) -> int:
p = 1 #工作节点
prefix = 1 #初始前缀，在树中移动
while(p < k):
count = self.getNum(prefix, n)
#count-1为除去根后该树节点数
if p + count-1 >= k: #在子树
prefix *= 10
p += 1 #字典树中父节点之后是其第一子
else: #在下一棵树
prefix += 1
p += count #字典树中结点的兄弟间的序号隔了一个子树
#print(p, prefix)
return prefix
def getNum(self, prefix, n): #获得prefix为根的树总结点数
cur = prefix
next = prefix + 1
count = 0
while(cur <= n):
count += min(next, n+1) - cur
cur *= 10
next *= 10
return count
```
---
# Agent Instructions: Querying This Documentation
If you need additional information that is not directly available in this page, you can query the documentation dynamically by asking a question.
Perform an HTTP GET request on the current page URL with the `ask` query parameter:
```
GET https://cai-sen-se.gitbook.io/leetcode/401-600/440.-zi-dian-xu-de-dikxiao-shu-zi.md?ask=
```
The question should be specific, self-contained, and written in natural language.
The response will contain a direct answer to the question and relevant excerpts and sources from the documentation.
Use this mechanism when the answer is not explicitly present in the current page, you need clarification or additional context, or you want to retrieve related documentation sections.