491. Non-decreasing Subsequences

排列用startindex

树枝不去重，树层去重

子集问题结果在结点（个数>=2）

class Solution(object):
    def findSubsequences(self, nums):
        """
        :type nums: List[int]
        :rtype: List[List[int]]
        """
        path = []
        result = []
        self.backtracking(nums, 0, path, result)
        return result


    def backtracking(self,nums, startIndex, path, result):
        if len(path) > 1:
            result.append(path[:])
        # 注意这里不要加return，要取树上的节点

        uset = set()
        for i in range(startIndex, len(nums)):
            if (path and nums[i]<path[-1]) or nums[i] in uset:
                continue
            uset.add(nums[i])
            path.append(nums[i])
            self.backtracking(nums,i+1, path, result)
            path.pop()
            

        

46. Permutations

排列问题不用startindex

因为排列问题，每次都要从头开始搜索，例如元素1在[1,2]中已经使用过了，但是在[2,1]中还要再使用一次1。

而used数组，其实就是记录此时path里都有哪些元素使用了，一个排列里一个元素只能使用一次。

class Solution(object):
    def permute(self, nums):
        """
        :type nums: List[int]
        :rtype: List[List[int]]
        """
        path =[]
        result = []
        used = [False]*len(nums)
        self.backtracking(nums,used, path, result )
        return result
    
    def backtracking(self, nums, used, path, result):
        if len(nums) == len(path):
            result.append(path[:])
            return #叶子节点收集

        for i in range(len(nums)):
            if used[i]:
                continue
            used[i]= True
            path.append(nums[i])
            self.backtracking(nums, used, path, result)
            path.pop()
            used[i]=False

47. Permutations II

树层去重（yes），树枝去重（no）

不用startindex

结合46和40

class Solution(object):
    def permuteUnique(self, nums):
        """
        :type nums: List[int]
        :rtype: List[List[int]]
        """
        path = []
        result = []
        nums.sort()
        used = [False] * len(nums)
        self.backtracking(nums, used, path, result)
        return result
    
    def backtracking(self, nums, used, path, result):
        if len(nums) == len(path):
            result.append(path[:])
            return

        for i in range(len(nums)):
            if i > 0 and nums[i] == nums[i-1] and not used[i-1]:
                continue
            if not used[i]:
                used[i] = True
                path.append(nums[i])
                self.backtracking(nums, used, path, result)
                path.pop()
                used[i] = False