题目：原题链接（中等）
标签：回溯算法
解法 | 时间复杂度 | 空间复杂度 | 执行用时 |
Ans 1 (Python) | O ( N × N ! ) | O ( N ) | 868ms (18.30%) |
Ans 2 (Python) | O ( N × N ! ) | O ( N ) | 680ms (18.69%) |
Ans 3 (Python) | O ( N × N ! ) | O ( N ) | 48ms (84.10%) |
Ans 4 (Python) | O ( N × N ! ) | O ( 1 ) | 64ms (37.83%) |
解法一（没有剪枝的回溯算法）：
class Solution:
def __init__(self):
self.visited = set()
self.ans = set()
self.now = []
def permuteUnique(self, nums: List[int]) -> List[List[int]]:
def track_back():
if len(self.now) == len(nums):
self.ans.add(tuple([nums[i] for i in self.now]))
for i in range(len(nums)):
if i not in self.visited:
self.visited.add(i)
self.now.append(i)
track_back()
self.now.pop()
self.visited.remove(i)
track_back()
return [list(e) for e in self.ans]解法二（直接向暂存数组中写入实际值）：
class Solution:
def __init__(self):
self.visited = set()
self.ans = set()
self.now = []
def permuteUnique(self, nums: List[int]) -> List[List[int]]:
def track_back():
if len(self.now) == len(nums):
self.ans.add(tuple(self.now[:]))
for i in range(len(nums)):
if i not in self.visited:
self.visited.add(i)
self.now.append(nums[i])
track_back()
self.now.pop()
self.visited.remove(i)
track_back()
return [list(e) for e in self.ans]解法三（增加剪枝条件）：
class Solution:
def __init__(self):
self.visited = set()
self.ans = []
self.now = []
def permuteUnique(self, nums: List[int]) -> List[List[int]]:
n = len(nums)
def track_back():
if len(self.now) == n:
self.ans.append(self.now[:])
tmp_set = set()
for i in range(n):
if i not in self.visited:
if nums[i] in tmp_set:
continue
tmp_set.add(nums[i])
self.visited.add(i)
self.now.append(nums[i])
track_back()
self.now.pop()
self.visited.remove(i)
track_back()
return self.ans解法四（不占用额外空间）：
class Solution:
def __init__(self):
self.nums = []
self.ans = []
def permuteUnique(self, nums: List[int]) -> List[List[int]]:
n = len(nums)
def track_back(now_idx):
if now_idx == n - 1:
self.ans.append(nums[:])
tmp_set = set()
for i in range(now_idx, n):
# 剪枝条件
if nums[i] in tmp_set:
continue
tmp_set.add(nums[i])
nums[i], nums[now_idx] = nums[now_idx], nums[i]
track_back(now_idx + 1)
nums[i], nums[now_idx] = nums[now_idx], nums[i]
track_back(0)
return self.ans
