Subsets

https://leetcode.com/problems/subsets/

思路1

这里其实就是把Combination拿到题目的思路2，收集那种组合树所有root到其余节点（不包括root本身）path。

class Solution:
    # @param S, a list of integer
    # @return a list of lists of integer

    def subsets(self, S):
        def dfs(depth, start, valuelist):
            res.append(valuelist)#回溯的时候 每一层都要记录为结果就行
            if depth == len(S): return
            for i in range(start, len(S)):
                dfs(depth+1, i+1, valuelist+[S[i]])#这里注意是使用的+一个list，即valuelist并没有被传入递归函数中，所以当这个函数返回之后，valuelist并没有变。例如，递归完左子树返回之后，valuelist依然还是递归左子树之前的值。这里传入的是合并的list，属于一个新list
                #所以这里其实是用python语言的特点，使得在模板中提到的改变变量->递归所有子节点->恢复变量的过程简化了
        S.sort()
        res = []
        dfs(0, 0, [])
        return res

思路2

类似与combination. 利用backtracking的模板

class Solution(object):

    def dfs(self, m, nums, subres, res):

        if m >= len(nums):
            tmp = [nums[i] for i,c in enumerate(subres) if c == 1]
            tmp.sort()
            res.append(tmp)
            return
        else:
            subres[m] = 0
            self.dfs(m+1, nums, subres, res)
            subres[m] = 1
            self.dfs(m+1, nums, subres, res)

    def subsets(self, nums):
        """
        :type nums: List[int]
        :rtype: List[List[int]]
        """
        subres = [0]*len(nums)
        res = []
        self.dfs(0, nums, subres, res)
        return res

思路2

还可以用bitmap的办法做
http://algorithm.yuanbin.me/zh-cn/exhaustive_search/subsets.html

自己重写code

class Solution(object):

    def dfs(self, candidates, start, end, subres, res):
        res.append(subres)
        for i in xrange(start, end):
            self.dfs(candidates, i + 1, end, subres + [candidates[i]], res)

    def subsets(self, nums):
        """
        :type nums: List[int]
        :rtype: List[List[int]]
        """
        nums.sort()
        res = []
        self.dfs(nums, 0, len(nums), [], res)
        return res

Subsets II

https://leetcode.com/problems/subsets-ii/

思路1

class Solution:
    # @param num, a list of integer
    # @return a list of lists of integer
    def subsetsWithDup(self, S):
        def dfs(depth, start, valuelist):
            if valuelist not in res: res.append(valuelist)
            if depth == len(S): return
            for i in range(start, len(S)):
                dfs(depth+1, i+1, valuelist+[S[i]])
        S.sort()
        res = []
        dfs(0, 0, [])
        return res

思路2

加上一个判断重复的条件即可

class Solution(object):

    def dfs(self, m, nums, subres, res):

        if m >= len(nums):
            tmp = [nums[i] for i,c in enumerate(subres) if c == 1]
            tmp.sort()
            if tmp not in res:#加上这个条件即可
                res.append(tmp)
            return
        else:
            subres[m] = 0
            self.dfs(m+1, nums, subres, res)
            subres[m] = 1
            self.dfs(m+1, nums, subres, res)


    def subsetsWithDup(self, nums):
        """
        :type nums: List[int]
        :rtype: List[List[int]]
        """
        subres = [0]*len(nums)
        res = []
        self.dfs(0, nums, subres, res)
        return res

自己重写code

class Solution(object):


    def dfs(self, candidates, start, end, subres, res):
        if subres not in res: res.append(subres)
        for i in xrange(start, end):
            self.dfs(candidates, i + 1, end, subres + [candidates[i]], res)

    def subsetsWithDup(self, nums):
        """
        :type nums: List[int]
        :rtype: List[List[int]]
        """
        nums.sort()
        res = []
        self.dfs(nums, 0, len(nums), [], res)
        return res