[LeetCode] 90. 子集 II

题目链接 : https://leetcode-cn.com/problems/subsets-ii/

题目描述:

给定一个可能包含重复元素的整数数组 nums,返回该数组所有可能的子集(幂集)。

说明:解集不能包含重复的子集。

示例:

输入: [1,2,2]
输出:
[
  [2],
  [1],
  [1,2,2],
  [2,2],
  [1,2],
  []
]

思路:

思路一:递归

思路二:迭代

直接看代码

代码:

思路一:

class Solution:
    def subsetsWithDup(self, nums: List[int]) -> List[List[int]]:
        res = []
        n = len(nums)
        nums.sort()
        def helper(idx, tmp):
            res.append(tmp)
            for i in range(idx, n):
                if i > idx and nums[i] == nums[i-1]:
                    continue
                helper(i+1, tmp + [nums[i]])
        helper(0, [])
        return res

java

class Solution {
    public List<List<Integer>> subsetsWithDup(int[] nums) {

        List<List<Integer>> res = new ArrayList<>();
        if (nums == null || nums.length == 0) return res;
        Arrays.sort(nums);
        backtrack(0, nums, res, new ArrayList<>());
        return res;


    }

    public void backtrack(int idx, int[] nums, List<List<Integer>> res, List<Integer> tmp_list) {
        res.add(new ArrayList<>(tmp_list));
        for (int i = idx; i < nums.length; i++) {
            if (i > idx && nums[i - 1] == nums[i]) continue;
            tmp_list.add(nums[i]);
            backtrack(i + 1, nums, res, tmp_list);
            tmp_list.remove(tmp_list.size() - 1);
        }
    }
}

思路二:

class Solution:
    def subsetsWithDup(self, nums: List[int]) -> List[List[int]]:
        if not nums: return []
        nums.sort()
        res = [[]]
        cur = []
        for i in range(len(nums)):
            if i > 0 and nums[i - 1] == nums[i]:
                cur = [tmp + [nums[i]] for tmp in cur]
            else:
                cur = [tmp + [nums[i]] for tmp in res]
            res += cur
        return res

类似题目还有:

39.组合总和

40. 组合总和 II

46. 全排列

47. 全排列 II

78. 子集

90. 子集 II

这类题目都是同一类型的,用回溯算法!

其实回溯算法关键在于:不合适就退回上一步

然后通过约束条件, 减少时间复杂度.

大家可以从下面的解法找出一点感觉!

78. 子集

class Solution:
	def subsets(self, nums):		
        if not nums:
			return []
		res = []
		n = len(nums)

		def helper(idx, temp_list):
			res.append(temp_list)
			for i in range(idx, n):
				helper(i + 1, temp_list + [nums[i]])

		helper(0, [])
		return res

90. 子集 II

class Solution(object):
    def subsetsWithDup(self, nums):
        """
        :type nums: List[int]
        :rtype: List[List[int]]
        """
        if not nums:
            return []
        n = len(nums)
        res = []
        nums.sort()
		# 思路1
        def helper1(idx, n, temp_list):
            if temp_list not in res:
                res.append(temp_list)
            for i in range(idx, n):
                helper1(i + 1, n, temp_list + [nums[i]])
		# 思路2
        def helper2(idx, n, temp_list):
            res.append(temp_list)
            for i in range(idx, n):
                if i > idx and  nums[i] == nums[i - 1]:
                    continue
                helper2(i + 1, n, temp_list + [nums[i]])

        helper2(0, n, [])
        return res

46. 全排列

class Solution(object):
    def permute(self, nums):
        """
        :type nums: List[int]
        :rtype: List[List[int]]
        """
        if not nums:
            return
        res = []
        n = len(nums)
        visited = [0] * n
        def helper1(temp_list,length):
            if length == n:
                res.append(temp_list)
            for i in range(n):
                if visited[i] :
                    continue
                visited[i] = 1
                helper1(temp_list+[nums[i]],length+1)
                visited[i] = 0
        def helper2(nums,temp_list,length):
            if length == n:
                res.append(temp_list)
            for i in range(len(nums)):
                helper2(nums[:i]+nums[i+1:],temp_list+[nums[i]],length+1)
        helper1([],0)
        return res

47. 全排列 II

class Solution(object):
    def permuteUnique(self, nums):
        """
        :type nums: List[int]
        :rtype: List[List[int]]
        """
        if not nums:
			return []
		nums.sort()
		n = len(nums)
		visited = [0] * n
		res = []

		def helper1(temp_list, length):
			# if length == n and temp_list not in res:
			# 	res.append(temp_list)
			if length == n:
				res.append(temp_list)
			for i in range(n):
				if visited[i] or (i > 0 and nums[i] == nums[i - 1] and not visited[i - 1]):
					continue
				visited[i] = 1
				helper1(temp_list + [nums[i]], length + 1)
				visited[i] = 0

		def helper2(nums, temp_list, length):
			if length == n and temp_list not in res:
				res.append(temp_list)
			for i in range(len(nums)):
				helper2(nums[:i] + nums[i + 1:], temp_list + [nums[i]], length + 1)

		helper1([],0)
		# helper2(nums, [], 0)
		return res

39.组合总和

class Solution(object):
    def combinationSum(self, candidates, target):
        """
        :type candidates: List[int]
        :type target: int
        :rtype: List[List[int]]
        """
        if not candidates:
            return []
        if min(candidates) > target:
            return []
        candidates.sort()
        res = []

        def helper(candidates, target, temp_list):
            if target == 0:
                res.append(temp_list)
            if target < 0:
                return
            for i in range(len(candidates)):
                if candidates[i] > target:
                    break
                helper(candidates[i:], target - candidates[i], temp_list + [candidates[i]])
        helper(candidates,target,[])
        return res

40. 组合总和 II

class Solution:
    def combinationSum2(self, candidates: List[int], target: int) -> List[List[int]]:
        if not candidates:
            return []
        candidates.sort()
        n = len(candidates)
        res = []
        
        def backtrack(i, tmp_sum, tmp_list):
            if tmp_sum == target:
                res.append(tmp_list)
                return 
            for j in range(i, n):
                if tmp_sum + candidates[j]  > target : break
                if j > i and candidates[j] == candidates[j-1]:continue
                backtrack(j + 1, tmp_sum + candidates[j], tmp_list + [candidates[j]])
        backtrack(0, 0, [])    
        return res

posted on 2019-06-15 15:51  威行天下  阅读(177)  评论(0编辑  收藏  举报

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