47. 全排列 II

给定一个可包含重复数字的序列 nums ，按任意顺序 返回所有不重复的全排列。

示例 1：

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

示例 2：

输入：nums = [1,2,3]
输出：[[1,2,3],[1,3,2],[2,1,3],[2,3,1],[3,1,2],[3,2,1]]

提示：

• 1 <= nums.length <= 8
• -10 <= nums[i] <= 10

1.回溯+暴力

class Solution:
    def permuteUnique(self, nums: List[int]) -> List[List[int]]:
        res = []
        flag = [True for i in range(len(nums))]
        def dfs(nums, temp):
            if len(temp) == len(nums):
                result = deepcopy(temp)
                if result not in res:
                    res.append(result)
            for i in range(len(nums)):
                if flag[i]:
                    temp.append(nums[i])
                    flag[i] = False
                    dfs(nums, temp)
                    temp.pop()
                    flag[i] = True
        dfs(nums, [])
        return res

2.排序+剪枝
Python:

class Solution:
    def permuteUnique(self, nums: List[int]) -> List[List[int]]:
        res = []
        flag = [True for i in range(len(nums))]
        nums.sort()
        def dfs(nums, temp, j):
            if j == len(nums):
                res.append(deepcopy(temp))
            for i in range(len(nums)):
                if not flag[i]:
                    continue
                if i > 0 and nums[i] == nums[i-1] and flag[i-1]:
                    continue
                temp.append(nums[i])
                flag[i] = False
                dfs(nums, temp, j+1)
                temp.pop()
                flag[i] = True
        dfs(nums, [], 0)
        return res

Java:

class Solution {
    public List<List<Integer>> permuteUnique(int[] nums) {
        List<List<Integer>> result = new ArrayList<List<Integer>>();
        int[] flag = new int[nums.length];
        Arrays.sort(nums);
        List<Integer> arr = new ArrayList<Integer>();
        backtrace(result, flag, nums, arr);
        return result;
    }

    public void backtrace(List<List<Integer>> result, int[] flag, int[] nums, List<Integer> b)
    {
        if(b.size() == nums.length)
        {
            result.add(new ArrayList<Integer>(b));
        }
        for(int i = 0; i < nums.length; i++)
        {
            if(flag[i] == 0)
            {
                if( i > 0 && nums[i-1] == nums[i] && flag[i-1] == 0)
                    continue;
                else
                {
                    flag[i] = 1;
                    b.add(nums[i]);
                    backtrace(result, flag, nums, b);
                    b.remove(b.size()-1);
                    flag[i] = 0;
                }
            }
        }
    }
}