和Permutation那道题目差不多，也是递归，不过要去重。先排序，再利用prevNum变量区分是否重复。

class Solution:
    # @param num, a list of integer
    # @return a list of lists of integers
    def permuteUnique(self, num):
        length = len(num)
        if length == 0: return []
        if length == 1: return [num]
        num.sort()
        res = []
        prevNum = None
        for curr in xrange(length):
            if num[curr] == prevNum: continue
            prevNum = num[curr]
            for j in self.permuteUnique(num[:curr] + num[curr + 1:]):
                res.append([num[curr]] + j)
        return res