47. Permutations II

这一题允许有重复。需要做的是排序，然后判断重复，

class Solution(object):
    def permuteUnique(self, nums):
        """
        :type nums: List[int]
        :rtype: List[List[int]]
        """
        visit = [0 for i in range(len(nums))]
        nums.sort()
        res = []
        self.bfs(visit, nums, res, [])
        return res
    def bfs(self, visit,nums, res, temp):
        if len(temp) == len(nums):
            res.append(temp)
            return
        for i in range(len(nums)):
            if visit[i] == 0:
                if i > 0 and visit[i-1] != 0 and nums[i-1] == nums[i]:
                    continue
                visit[i] = 1
                self.bfs(visit, nums, res, temp+[nums[i]])
                visit[i] = 0