[Lintcode]16. Permutations II/[Leetcode]47. Permutations II

16. Permutations II/47. Permutations II

• 本题难度: Medium
• Topic: Search & Recursion

Description

Given a list of numbers with duplicate number in it. Find all unique permutations.

Example
Example 1:

Input: [1,1]
Output:
[
[1,1]
]
Example 2:

Input: [1,2,2]
Output:
[
[1,2,2],
[2,1,2],
[2,2,1]
]
Challenge
Using recursion to do it is acceptable. If you can do it without recursion, that would be great!

我的代码

法1: 递归 BFS

class Solution:
    """
    @param: :  A list of integers
    @return: A list of unique permutations
    """
    
    #recursion
    #dfs
    def permuteUnique(self, nums):
        # write your code here
        nums.sort()
        res = []
        n = len(nums)
        self.dfs(nums,[],res,n)
        return res
    
    def dfs(self,nums,path,res,n):
        if len(path) == n:
            res.append(path)
        for i in range(len(nums)):
            if i>0 and nums[i-1]==nums[i]:
                continue
            self.dfs(nums[:i]+nums[i+1:],path+[nums[i]],res,n)
            

法2 非递归 DFS

    #no recursion
    def permuteUnique(self, nums):
        # write your code here
        res = []
        nums.sort()
        containt = [[[], nums]]
        while (len(containt) > 0):
            [l1, l2] = containt.pop()
            if len(l2) == 0:
                res.append(l1)
            for i in range(len(l2)):
                if i > 0 and l2[i - 1] == l2[i]:
                    continue
                containt.append([l1 + [l2[i]], l2[:i] + l2[i + 1:]])
    
        return res

法3: 非递归，BFS

#BFS
def permuteUnique(nums):
    # write your code here
    res = []
    nums.sort()
    containt = collections.deque([[[], nums]])
    while (len(containt) > 0):
        [l1, l2] = containt.popleft()
        print('l1',l1,'l2',l2)
        if len(l2) == 0:
            res.append(l1)
        for i in range(len(l2)):
            if i > 0 and l2[i - 1] == l2[i]:
                continue
            containt.append([l1 + [l2[i]], l2[:i] + l2[i + 1:]])

    return res

别人的代码

均为非递归，但是没看懂，以后再看

1

       
    def permuteUnique(self, nums):
        ans = [[]]
        for n in nums:
            new_ans = []
            for l in ans:
                for i in range(len(l)+1):
                    new_ans.append(l[:i]+[n]+l[i:])
                    if i<len(l) and l[i]==n: break              #handles duplication
            ans = new_ans
        return ans

2

    def permute(nums):
        permutations = [[]]
        
        for head in nums:
            permutations = [rest[:i]+[head]+rest[i:] for rest in permutations for i in range(len(rest)+1)]
            
        return permutations

思路

递归比较好理解，DFS。
用栈是DFS
用队列是BFS

• 时间复杂度 O(n!)