47. Permutations II (全排列有重复的元素)

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],
  [2,1,1]
]


与上一题不同，就是在19行加个判断即可。

class Solution(object):
    def __init__(self):
        self.res = []

    def permuteUnique(self, nums):
        """
        :type nums: List[int]
        :rtype: List[List[int]]
        """
        self.help(nums, 0, len(nums))

        return self.res

    def help(self, a, lo, hi):
        if(lo == hi):
            self.res.append(a[0:hi])
        for i in range(lo, hi):
            #判断 i 是否已经在当过头元素了
            if a[i] not in a[lo:i]:
                self.swap(a, i, lo)
                self.help(a, lo + 1, hi)
                self.swap(a, i, lo)
    def swap(self, a, i, j):
        temp = a[i]
        a[i] = a[j]
        a[j] = temp