[LeetCode] 77. Combinations_Medium tag: backtracking

Given two integers n and k, return all possible combinations of k numbers out of the range [1, n].

You may return the answer in any order.

 

Example 1:

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

Example 2:

Input: n = 1, k = 1
Output: [[1]]

 

Constraints:

• 1 <= n <= 20
• 1 <= k <= n

 

Update on 10/30/2023, we use permutation 模板，然后相当于判断 nums[i] > temp[-1] 即可，然后当len(temp) == k时，ans.append(temp) 

Ideas:

1. 利用backtracking

T: k * C(k, n)  => k * n! / k! / (n - k)!  

S:  k * C(k, n)  => k * n! / k! / (n - k)!  

Code:

class Solution:
    def combine(self, n: int, k: int) -> List[List[int]]:
        nums = [i for i in range(1, n + 1)]

        ans = []
        self.helper(nums, [], ans, k)
        return ans
    
    def helper(self, nums, temp, ans, k):
        if len(temp) == k:
            ans.append(temp)
        elif len(temp) < k:
            for i in range(len(nums)):
                if not temp or nums[i] > temp[-1]:
                    self.helper(nums[:i] + nums[i + 1:], temp + [nums[i]], ans, k)
        

 

 

2. 利用subset模板，相当于subset只取len(temp) == k的情况：

class Solution:
    def combine(self, n: int, k: int) -> List[List[int]]:
        nums = [i for i in range(1, n + 1)]

        ans = []
        self.helper(nums, [], ans, 0, k)
        return ans
    
    def helper(self, nums, temp, ans, pos, k):
        if len(temp) == k:
            ans.append(temp)
        elif len(temp) < k:
            for i in range(pos, len(nums)):
                self.helper(nums, temp + [nums[i]], ans, i + 1, k)

 

3. 利用itertools.combinations

T: k * C(k, n)  => k * n! / k! / (n - k)!  

S:  k * C(k, n)  => k * n! / k! / (n - k)!  

Code:

class Solution:
    def combine(self, n: int, k: int) -> List[List[int]]:
        nums = [i for i in range(1, n + 1)]
        return itertools.combinations(nums, k)