Datawhale编程——递归

因为刚考完试不久,时间实在太赶了,就看了下题,思路还不是很清晰,第一题还算是自己做的,第二题就只能看答案了。在此对认真点评我的小伙伴说声抱歉,最后一次一定认真做。

leetcode 017

代码实现

class Solution:
    def letterCombinations(self, digits):
        """
        :type digits: str
        :rtype: List[str]
        """
        mapping = {'2': 'abc', '3': 'def', '4': 'ghi', '5': 'jkl', 
                   '6': 'mno', '7': 'pqrs', '8': 'tuv', '9': 'wxyz'}
        if len(digits) == 0:
            return []
        if len(digits) == 1:
            return list(mapping[digits[0]])
        prev = self.letterCombinations(digits[:-1])
        additional = mapping[digits[-1]]
        return [s + c for s in prev for c in additional]

leetcode 046

代码实现

class Solution:
    
    def dfs(self, nums, path, res):
        """
        :type nums: List[int]
        :type path: int
        :type res: List[List[int]]
        """
        if not nums:
            res.append(path)
            # return # backtracking
        for i in range(len(nums)):
            self.dfs(nums[:i]+nums[i+1:], path+[nums[i]], res)
            
    def permute(self, nums):
        """
        :type nums: List[int]
        :rtype: List[List[int]]
        """
        res = []
        self.dfs(nums, [], res)
        return res

原文地址:https://www.cnblogs.com/ChanWunsam/p/10241748.html