算法第25天 | 491. 非递减子序列、46. 全排列、47. 全排列 II

491. 非递减子序列

题目

思路与解法

class Solution:def findSubsequences(self, nums: List[int]) -> List[List[int]]:self.res = []self.path = []self.backtracking(nums, 0)return self.resdef backtracking(self, nums, starIdx:int):if len(self.path) >= 2:self.res.append(self.path[:])i = starIdxused = set()while i < len(nums):if (not self.path or nums[i] >= self.path[-1]) and nums[i] not in used:used.add(nums[i])self.path.append(nums[i])self.backtracking(nums, i+1)self.path.pop()i += 1

46. 全排列

题目

思路与解法

class Solution:def permute(self, nums: List[int]) -> List[List[int]]:self.res = []self.path = []self.backtracking(nums)return self.resdef backtracking(self, nums):if len(self.path) == len(nums):self.res.append(self.path[:])returnfor i in range(len(nums)):if nums[i] not in self.path:self.path.append(nums[i])self.backtracking(nums)self.path.pop()

47. 全排列 II

题目

思路与解法

class Solution:def permuteUnique(self, nums):nums.sort()  # 排序result = []self.backtracking(nums, [], [False] * len(nums), result)return resultdef backtracking(self, nums, path, used, result):if len(path) == len(nums):result.append(path[:])returnfor i in range(len(nums)):if (i > 0 and nums[i] == nums[i - 1] and not used[i - 1]) or used[i]: # used[i-1] 为false，表明本层使用过。为True，表示是别的层使用的，那么相同的值这一层就可以用continueused[i] = Truepath.append(nums[i])self.backtracking(nums, path, used, result)path.pop()used[i] = False