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 N ) | 240ms (33.07%) |
Ans 2 (Python) | O ( N N ) | O ( N N ) | 44ms (87.12%) |
Ans 3 (Python) |
解法一（暴力算法）：
class Solution:
def permutation(self, S: str) -> List[str]:
if len(S) == 1:
return [S]
elif len(S) == 2:
if S[0] == S[1]:
return [S]
else:
return [S, S[::-1]]
else:
ans = set()
for i in range(len(S)):
ch = S[i]
for other in self.permutation(S[:i] + S[i + 1:]):
ans.add(ch + other)
return list(ans)
解法二（更好的暴力）：
class Solution:
def permutation(self, S: str) -> List[str]:
if len(S) == 1:
return [S]
elif len(S) == 2:
if S[0] == S[1]:
return [S]
else:
return [S, S[::-1]]
else:
ans = set()
visited = set()
for i in range(len(S)):
ch = S[i]
if ch not in visited:
visited.add(ch)
for other in self.permutation(S[:i] + S[i + 1:]):
ans.add(ch + other)
return list(ans)
