题目：原题链接（中等）
标签：回溯算法、哈希表
解法 | 时间复杂度 | 空间复杂度 | 执行用时 |
Ans 1 (Python) | O ( 2 N ) | O ( 2 N ) | 40ms (69.07%) |
Ans 2 (Python) | |||
Ans 3 (Python) |
解法一：
# ---------- 生成所有不同的全排列(0047题) ----------
def permuteUnique(nums):
visited = set()
ans = []
now = []
n = len(nums)
def track_back():
if len(now) == n:
ans.append(now[:])
tmp_set = set()
for i in range(n):
if i not in visited:
if nums[i] in tmp_set:
continue
tmp_set.add(nums[i])
visited.add(i)
now.append(nums[i])
track_back()
now.pop()
visited.remove(i)
track_back()
return ans
class Solution:
def generatePalindromes(self, s: str) -> List[str]:
count = collections.Counter(s)
once = []
twice = []
for k, v in count.items():
n, m = divmod(v, 2)
if n:
twice.extend([k] * n)
if m:
once.append(k)
if len(once) > 1:
return []
ans = []
for lst in permuteUnique(twice):
ans.append("".join(lst + once + list(reversed(lst))))
return ans
