Given a collection of numbers, return all possible permutations.
For example,
[1,2,3]
have the following permutations:
[1,2,3]
, [1,3,2]
, [2,1,3]
, [2,3,1]
, [3,1,2]
,
and [3,2,1]
.
This problem is exponential. We can use one O(n) extra space to mark whether each element is used or not.
It is DFS.
class Solution:
# @param num, a list of integer
# @return a list of lists of integers
def bfs(self,num,val,res,used):
if len(val)==len(num) and val not in res:
res.append(val)
for i in range(len(num)):
if used[i]==False:
used[i]=True
val=val+[num[i]]
self.bfs(num,val,res,used)
val=val[:len(val)-1]
used[i]=False
def permute(self, num):
res=[]
if len(num)==0:
return res
used=[False]*len(num)
self.bfs(num,[],res,used)
return res