dfs全排列总结

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 nums are 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()