【leetcode题解】【回溯】【54】【M】Subsets II

Given a collection of integers that might contain duplicates, nums, return all possible subsets.

Note:

Elements in a subset must be in non-descending order.
The solution set must not contain duplicate subsets.

For example,

If nums =

[1,2,2]

, a solution is:

[
[2],
[1],
[1,2,2],
[2,2],
[1,2],
[]
]

Subscribe to see which companies asked this question

方法一，跳过相同元素（不能用if，得用while，总忘）
方法二，跟subset一样，添加结果的时候判断，是否重复

class Solution(object):
def bt(self,nums,temp,start,l,res):
#print start,l,temp
if len(temp) > l:
return
if len(temp) == l:# and temp not in res:
#print 'added'
res.append(temp[:])
return

i = start
while i <len(nums):
#temp.append(nums[i])
self.bt(nums,temp+[nums[i]],i+1,l,res)

while i<len(nums)-1 and nums[i] == nums[i+1]:
i+=1
continue
i += 1
#temp.pop(-1)

def subsetsWithDup(self, nums):
nums.sort()
res = []

for i in range(0,len(nums)+1):
self.bt(nums,[],0,i,res)

return res

'''
class Solution(object):
def bt(self,nums,temp,start,l,res):
#print start,l,temp
if len(temp) > l:
return
if len(temp) == l and temp not in res:
res.append(temp[:])
#return
for i in range(start,len(nums)):
#temp.append(nums[i])

self.bt(nums,temp+[nums[i]],i+1,l,res) #最开始i+1 写成了 start+1，那肯定是错了啊！！
#temp.pop(-1)

def subsetsWithDup(self, nums):

res = []
nums.sort()

for i in range(0,len(nums)+1):
#print '______'
self.bt(nums,[],0,i,res)
return res
'''