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64 lines
1.7 KiB
Plaintext
64 lines
1.7 KiB
Plaintext
---
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title: '46. Permutations'
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description: Given an array nums of distinct integers, return all the possible permutations. You can return the answer in any order
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icon: dot
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topics:
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- { name: "Array", slug: "array" }
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- { name: "Backtracking", slug: "backtracking" }
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issue: "https://github.com/prdlk/leetcode/issues/151"
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companies:
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- { name: "Bloomberg", period: "Thirty Days", frequency: 69.8 }
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- { name: "Amazon", period: "Thirty Days", frequency: 48.9 }
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- { name: "Microsoft", period: "Three Months", frequency: 42.1 }
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- { name: "Google", period: "Three Months", frequency: 32.7 }
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- { name: "Infosys", period: "Six Months", frequency: 67.9 }
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- { name: "LinkedIn", period: "Six Months", frequency: 54.1 }
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- { name: "Meta", period: "Six Months", frequency: 43.4 }
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sidebar:
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label: 'Permutations'
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badge: 'Medium'
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---
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### Example 1:
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- Input: `nums = [1,2,3]`
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- Output: `[[1,2,3],[1,3,2],[2,1,3],[2,3,1],[3,1,2],[3,2,1]]`
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### Example 2:
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- Input: `nums = [0,1]`
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- Output: `[[0,1],[1,0]]`
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### Example 3:
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- Input: `nums = [1]`
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- Output: `[[1]]`
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### Constraints:
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- `1 <= nums.length <= 6`
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- `-10 <= nums[i] <= 10`
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- All the integers of `nums` are unique.
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## Solution
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```py
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class Solution:
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def permute(self, nums: list[int]) -> list[list[int]]:
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result, path = [], []
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used = [False] * len(nums)
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def backtrack():
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if len(path) == len(nums):
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result.append(path[:])
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return
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for i in range(len(nums)):
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if used[i]:
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continue
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used[i] = True
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path.append(nums[i])
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backtrack()
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path.pop()
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used[i] = False
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backtrack()
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return result
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```
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