90. Subsets II Leetcode Python

最新推荐文章于 2021-04-08 16:46:45 发布

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本文介绍了一种处理包含重复元素集合的所有可能子集的算法。通过递归方式结合循环来避免生成重复的子集，并确保子集元素按非递减顺序排列。讨论了时间复杂度为NP-hard的问题，并提供了Python实现代码。

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Given a collection of integers that might contain duplicates, S, return all possible subsets.

Note:

Elements in a subset must be in non-descending order.
The solution set must not contain duplicate subsets.

For example,
If S = [1,2,2], a solution is:

[
  [2],
  [1],
  [1,2,2],
  [2,2],
  [1,2],
  []
]

算法NP hard T(n) = n * T(n-1)

这道题目的解法和大部分的求combination和subset的方法一样，在递归中加一个循环逐次加入新的元素。

这道题目看似O(n^2)起始不然，随着数组的增大子集的数量是factorial 增长的。

This problem can be solve with adding one iteration in recursion. and append the subsets to the solution list. The time complexity is NP hard, because the number of sub array is factorial related with the size of the array.

class Solution:
    # @param num, a list of integer
    # @return a list of lists of integer
    def bfs(self,valuelist,solution,S,start):
        if valuelist not in solution and len(valuelist)<=len(S):
            solution.append(valuelist)
        for index in range(start,len(S)):
            valuelist=valuelist+[S[index]]
            self.bfs(valuelist,solution,S,index+1)
            valuelist=valuelist[:len(valuelist)-1]
    def subsetsWithDup(self, S):
        solution=[]
        if len(S)==0:
            return solution
        S.sort()
        self.bfs([],solution,S,0)
        return solution

Simplified version

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