些許時間沒更新blog了XD
這次放LeetCode來分享好了
https://leetcode.com/problems/permutations-ii/
其實這題延續第一題
https://leetcode.com/problems/permutations/
第一題用DFS做就好了
那第二題其實也不用改太多，在同一個node下，跳過相同value的subnode就好了
|
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 |
class Solution(object): def permuteUnique(self, nums): """ :type nums: List[int] :rtype: List[List[int]] """ res=[] nums.sort() self.dfs(nums, res, [], len(nums)) return res def dfs(self, nums, res, current,length): if not nums: res.append(current) else: prev = nums[0] for i in range(length): if prev == nums[i] and i != 0: continue else: self.dfs(nums[:i]+nums[i+1:], res, current+[nums[i]],length-1) prev = nums[i] |
用python寫可以跳過整理array的部分比C方便多惹….
但複雜度應該沒甚麼影響都一樣慢XD
