[LeetCode] 47. 全排列 II

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

题目描述:

给定一个可包含重复数字的序列，返回所有不重复的全排列。

示例:

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

思路:

思路一: 库函数

class Solution:
    def permuteUnique(self, nums: List[int]) -> List[List[int]]:
        return list({p for p in itertools.permutations(nums)})

思路二: 回溯算法

通过排序,来减少重复数组进入res

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代码:

class Solution:
    def permuteUnique(self, nums: List[int]) -> List[List[int]]:
        nums.sort()
        res = []
        visited = set()
        def backtrack(nums, tmp):
            if len(nums) == len(tmp):
                res.append(tmp)
                return
            for i in range(len(nums)):
                if i in visited or (i > 0 and i - 1 not in visited and nums[i-1] == nums[i]):
                    continue
                visited.add(i)
                backtrack(nums, tmp + [nums[i]])
                visited.remove(i)
        backtrack(nums, [])
        return res

class Solution {
    public List<List<Integer>> permuteUnique(int[] nums) {
        List<List<Integer>> res = new ArrayList<>();
        Arrays.sort(nums);
        int[] visited = new int[nums.length];
        backtrack(res, nums, visited, new ArrayList<Integer>());
        return res;

    }

    private void backtrack(List<List<Integer>> res, int[] nums, int[] visited, ArrayList<Integer> tmp) {
        if (tmp.size() == nums.length) {
            res.add(new ArrayList<>(tmp));
            return;
        }
        for (int i = 0; i < nums.length; i++) {
            if (visited[i] == 1 || (i > 0 && visited[i - 1] == 0 && nums[i - 1] == nums[i])) continue;
            visited[i] = 1;
            tmp.add(nums[i]);
            backtrack(res, nums, visited, tmp);
            tmp.remove(tmp.size() - 1);
            visited[i] = 0;
        }
    }
}

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