17. Letter Combinations of a Phone Number
Medium
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Given a string containing digits from 2-9 inclusive, return all possible letter combinations that the number Could represent. Return the answer in any order.
A mapping of digits to letters (just like on the telephone buttons) is given below. Note that 1 does not map to any letters.
Example 1:
Input: digits = "23" Output: ["ad","ae","af","bd","be","bf","cd","ce","cf"]
Example 2:
Input: digits = "" Output: []
Example 3:
Input: digits = "2" Output: ["a","b","c"]
Constraints:
0 <= digits.length <= 4-
digits[i]is a digit in the range['2', '9'].
Accepted
1.3M
Submissions
2.4M
遇到这样的题首先可以画图。全排列分为两种，2叉树和k叉树，这题就是个k叉树。
比如输入“23”，即可画出下图。
由此我们可以把他转换成dfs问题，只要把k叉树遍历即可。
首先考虑遍历的终止条件，就是每个digit都访问到了的情况，所以我们使用index来记录目前访问了几位digit
在访问时用cur 字符串来记录当前遍历路径所生成的字符串
class Solution:
def letterCombinations(self, digits: str) -> List[str]:
if not digits:
return None
self.res = []
self.hm = {"2":"abc","3":"def",
"4":"ghi","5":"jkl","6":"mno"
,"7":"pqrs","8":"tuv","9":"wxyz"}
self.dfs(digits,"",0)
return self.res
def dfs(self,digits,cur,index):
if index == len(digits):
self.res.append(cur)
return
for c in self.hm[digits[index]]:
self.dfs(digits,cur+c,index+1)
78. Subsets
Medium
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Given an integer array nums of unique elements, return all possible subsets (the power set).
The solution set must not contain duplicate subsets. Return the solution in any order.
Example 1:
Input: nums = [1,2,3] Output: [[],[1],[2],[1,2],[3],[1,3],[2,3],[1,2,3]]
Example 2:
Input: nums = [0] Output: [[],[0]]
Constraints:
1 <= nums.length <= 10-10 <= nums[i] <= 10- All the numbers of
numsare unique.
这个题是一个二叉树问题，对于列表里的每个元素，我们都可以选择加入或不加入列表。
使用dfs 即可，因为有回溯所以每次加完了需要pop
class Solution:
def subsets(self, nums: List[int]) -> List[List[int]]:
self.res = []
self.dfs(nums,0,[])
return self.res
def dfs(self,nums,index,cur):
if index == len(nums):
self.res.append(list(cur))
return
self.dfs(nums, index + 1, cur)
cur.append(nums[index])
self.dfs(nums, index + 1, cur)
cur.pop()
