[leetcode] 77. Combinations @ python

原题

Given two integers n and k, return all possible combinations of k numbers out of 1 … n.

Example:

Input: n = 4, k = 2
Output:
[
[2,4],
[3,4],
[2,3],
[1,2],
[1,3],
[1,4],
]

解法

DFS+回朔法. DFS的关键是找到回溯的条件, 当k= 0时, 表示我们已经取完数字, 此时将path加到结果中, 然后加限制条件: nums的长度需要大于等于K才能取完数字, 然后遍历nums, 递归寻找结果.

Time: O(n)
Space: O(1)

代码

class Solution(object):def combine(self, n, k):""":type n: int:type k: int:rtype: List[List[int]]"""def dfs(nums, k, path, res):if k == 0:res.append(path)returnif len(nums) >= k:for i in range(len(nums)+1-k):dfs(nums[i+1:], k-1, path + [nums[i]], res)res = []nums = list(range(1, n+1))dfs(nums, k, [], res)return res

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[leetcode] 77. Combinations @ python

原题

解法

代码

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