给定一个可包含重复数字的序列 nums ,按任意顺序 返回所有不重复的全排列。
示例 1:
输入:nums = [1,1,2]
输出:
[[1,1,2],
[1,2,1],
[2,1,1]]
示例 2:
输入:nums = [1,2,3]
输出:[[1,2,3],[1,3,2],[2,1,3],[2,3,1],[3,1,2],[3,2,1]]
提示:
1 <= nums.length <= 8
-10 <= nums[i] <= 10
class Solution:
def __init__(self):
self.res = []
self.track = []
self.used = []
def permuteUnique(self, nums: List[int]) -> List[List[int]]:
# 先排序,让相同的元素靠在一起
nums.sort()
self.used = [False] * len(nums)
self.backtrack(nums)
return self.res
def backtrack(self, nums: List[int]) -> None:
if len(self.track) == len(nums):
self.res.append(self.track[:])
return
for i in range(len(nums)):
if self.used[i]:
continue
# 新添加的剪枝逻辑,固定相同的元素在排列中的相对位置
if i > 0 and nums[i] == nums[i - 1] and not self.used[i - 1]:
continue
self.track.append(nums[i])
self.used[i] = True
self.backtrack(nums)
self.track.pop()
self.used[i] = False