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