代码随想录算法训练营day31 | 491.递增子序列、46.全排列、47.全排列 II

491.递增子序列

未去重的代码

class Solution:def findSubsequences(self, nums: List[int]) -> List[List[int]]:result = []self.backtracking(nums, result, [], 0)return resultdef backtracking(self, nums, result, path, startIndex):if len(path) >= 2:result.append(path[:])for i in range(startIndex, len(nums)):if len(path) >= 1 and nums[i] < path[-1]:continuepath.append(nums[i])self.backtracking(nums, result, path, i+1)path.pop()

本题是树结构的同一层剪枝

但是本题不能排序，和之前的剪枝差别很大

class Solution:def findSubsequences(self, nums: List[int]) -> List[List[int]]:result = []self.backtracking(nums, result, [], 0)return resultdef backtracking(self, nums, result, path, startIndex):if len(path) >= 2:result.append(path[:])used = set()for i in range(startIndex, len(nums)):if len(path) >= 1 and nums[i] < path[-1]:continueif nums[i] in used:continueused.add(nums[i])path.append(nums[i])self.backtracking(nums, result, path, i+1)path.pop()

优化点：因为给定的数字范围是-100 <= nums[i] <= 100，因此set可以优化为数组

46.全排列

class Solution:def permute(self, nums: List[int]) -> List[List[int]]:result = []used = [False] * len(nums)self.backtracking(nums, result, [], used)return resultdef backtracking(self, nums, result, path, used):if len(path) == len(nums):result.append(path[:])returnfor i in range(len(nums)):if used[i]:continuepath.append(nums[i])used[i] = Trueself.backtracking(nums, result, path, used)path.pop()used[i] = False

47.全排列 II

去重的思路需要继续理解

class Solution:def permuteUnique(self, nums: List[int]) -> List[List[int]]:result = []used = [False] * len(nums)nums.sort()self.backtracking(nums, result, [], used)return resultdef backtracking(self, nums, result, path, used):if len(path) == len(nums):result.append(path[:])returnfor i in range(len(nums)):if used[i]:continueif i > 0 and nums[i] == nums[i-1] and not used[i-1]:continuepath.append(nums[i])used[i] = Trueself.backtracking(nums, result, path, used)path.pop()used[i] = False