[Leetcode]Permutations II

Given a collection of numbers that might contain duplicates, return all possible unique permutations.

For example,
[1,1,2] have the following unique permutations:

[1,1,2], [1,2,1], and [2,1,1].

 Permutations的扩展题，不同的是这题里可能会出现重复元素~解法其实也是类似的，注意处理重复元素就可以了~ 先对数组排序，然后判断当前元素是否与上一个元素相同~

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

class Solution:    # @param num, a list of integer    # @return a list of lists of integers    def permuteUnique(self, num):        if num is None or len(num) == 0: return []        if len(num) == 1: return [num]        num.sort()        self.res = []        self.helper(num, [False] * len(num), [])        return self.res            def helper(self, num, used, item):        if len(item) == len(num):            self.res.append(item)            return        for i in xrange(len(num)):            if i > 0 and not used[i - 1] and num[i] == num[i - 1]: continue            if not used[i]:                used[i] = True                self.helper(num, used, item + [num[i]])                used[i] = False