47. 全排列 II(Permutations II)M

英文题目

Given a collection of numbers, nums, that might contain duplicates, return all possible unique permutations in any order.
Example 1:

Input: nums = [1,1,2]
Output:
[[1,1,2],
 [1,2,1],
 [2,1,1]]

Example 2:

Input: nums = [1,2,3]
Output: [[1,2,3],[1,3,2],[2,1,3],[2,3,1],[3,1,2],[3,2,1]]

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

中文题目

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

回溯

时间复杂度 O(n x n!)，空间复杂度O(n)

# python3: 时间 44 ms, 击败 89.75%; 内存 16.1 MB, 击败 67.5
class Solution:
    def permuteUnique(self, nums: List[int]) -> List[List[int]]:
        self.res = []
        used = [False] * len(nums)
        nums.sort()
        self.backtrack(nums, [], used)
        return self.res
    
    def backtrack(self, nums: List[int], path: List[int], 
                  used: List[bool]) -> None:
        if len(nums) == len(path):
            self.res.append(path[:])
            return
        
        for i in range(len(nums)):
            if used[i]:
                continue
            elif i > 0 and nums[i] == nums[i-1] and used[i-1] == False:
                continue
            used[i] = True
            path.append(nums[i])
            self.backtrack(nums, path, used)
            path.pop()
            used[i] = False

// c++: 时间 8 ms, 击败 61.48%; 内存 8.3 MB, 击败 87.84%
class Solution {
public:
    vector<vector<int>> permuteUnique(vector<int>& nums) {
        vector<vector<int>> res;
        vector<int> path;
        vector<bool> used(nums.size(), false);
        sort(nums.begin(), nums.end());
        backtrack(res, nums, path, used);
        return res;
    }

    void backtrack(vector<vector<int>> &res, const vector<int> &nums,
                   vector<int> &path, vector<bool> &used) {
        if (nums.size() == path.size()) {
            res.push_back(path);
            return;
        }

        for (int i = 0; i < nums.size(); ++i) {
            if (used[i]) {
                continue;
            } else if (i > 0 && nums[i] == nums[i-1] && used[i-1] == false) {
                continue;
            }
            used[i] = true;
            path.push_back(nums[i]);
            backtrack(res, nums, path, used);
            path.pop_back();
            used[i] = false;
        }
    }
};

// java: 时间 1 ms, 击败 99.65%; 内存 42.7 MB, 击败 54.66%
class Solution {
    public List<List<Integer>> permuteUnique(int[] nums) {
        List<List<Integer>> res = new ArrayList<>();
        List<Integer> path = new ArrayList<>();
        boolean[] used = new boolean[nums.length];
        Arrays.fill(used, false);
        Arrays.sort(nums);
        backtrack(res, nums, path, used);
        return res;
    }

    private void backtrack(List<List<Integer>> res, int[] nums,
                           List<Integer> path, boolean[] used) {
        if (path.size() == nums.length) {
            // 注意结果要 new 一个新arraylist
            res.add(new ArrayList<Integer>(path));
            return;
        }

        for (int i = 0; i < nums.length; ++i) {
            if (used[i]) {
                continue;
            } else if (i > 0 && nums[i] == nums[i-1] && used[i-1] == false) {
                continue;
            }
            used[i] = true;
            path.add(nums[i]);
            backtrack(res, nums, path, used);
            path.remove(path.size()-1);
            used[i] = false;
        }
    }
}

// go: 时间 0 ms, 击败 100%; 内存 3.5 MB, 击败 95.14%
func permuteUnique(nums []int) [][]int {
    res := make([][]int, 0)
    path := make([]int, 0, len(nums))
    used := make([]bool, len(nums))
    sort.Ints(nums)
    backtrack(&res, nums, path, used)
    return res
}

func backtrack(res *[][]int, nums []int, path []int, used []bool) {
    if len(nums) == len(path) {
        tmp := make([]int, len(nums))
        copy(tmp, path)
        *res = append(*res, tmp)
    }
    for i := 0; i < len(nums); i++ {
        if used[i] {
            continue
        } else if i > 0 && nums[i] == nums[i-1] && used[i-1] == false {
            continue
        }
        used[i] = true
        path = append(path, nums[i])
        backtrack(res, nums, path, used)
        path = path[:len(path)-1]
        used[i] = false
    }
}

// javascript: 时间 64 ms, 击败 97.68%; 内存 44 MB, 击败 67.18%
/**
 * @param {number[]} nums
 * @return {number[][]}
 */
var permuteUnique = function(nums) {
    const res = new Array();
    const path = new Array();
    const used = new Array(nums.length).fill(false);
    nums.sort((a,b) => a-b);
    backtrack(res, nums, path, used);
    return res;
};

const backtrack = function(res, nums, path, used) {
    if (nums.length == path.length) {
        res.push(Array.from(path));
        return;
    }

    for (let i = 0; i < nums.length; ++i) {
        if (used[i]) {
            continue;
        } else if (i > 0 && nums[i] == nums[i-1] && used[i-1] == false) {
            continue;
        }
        used[i] = true;
        path.push(nums[i]);
        backtrack(res, nums, path, used);
        path.pop();
        used[i] = false;
    }
}

# 47. 全排列 II(Permutations II)M

# 英文题目

# 中文题目

# 回溯

47. 全排列 II(Permutations II)M

英文题目

中文题目

回溯