46. Permutations

1. Description

Given an array nums of distinct integers, return all the possible permutations. You can return the answer in any order.

2. Example

Example 1

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

Example 2

Input: nums = [0,1]
Output: [[0,1],[1,0]]

Example 3

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

3. Constraints

  • 1 <= nums.length <= 6
  • -10 <= nums[i] <= 10
  • All the integers of nums are unique.

4. Solutions

Backtracking

n = nums.size()
Time complexity: O(nn!)
Space complexity: O(n)

class Solution {
public:
    vector<vector<int>> permute(vector<int> nums) {
        vector<vector<int>> permutations;
        search_permutations(permutations, nums, 0);

        return permutations;
    }

private:
    void search_permutations(
        vector<vector<int>> &permutations,
        vector<int> &permutation,
        int index) {
        const int n = permutation.size();
        if (index == n) {
            permutations.emplace_back(permutation);
        } else {
            for (int i = index; i < n; ++i) {
                swap(permutation[i], permutation[index]);
                search_permutations(permutations, permutation, index + 1);
                swap(permutation[i], permutation[index]);
            }
        }
    }
};
Permutation

n = nums.size()
Time complexity: O(nn!)
Space complexity: O(n)

class Solution {
public:
    vector<vector<int>> permute(const vector<int> &nums) {
        const int n = nums.size();
        int count = 1;
        for (int i = n; i > 0; --i) {
            count *= i;
        }

        vector<vector<int>> combinations;
        combinations.reserve(count);
        combinations.push_back(nums);

        for (int i = 1; i < count; ++i) {
            combinations.emplace_back(next_permutation(combinations[i - 1]));
        }

        return combinations;
    }

private:
    vector<int> next_permutation(vector<int> nums) {
        const int n = nums.size();
        int last_inc_index = n - 1;
        while (last_inc_index > 0 && nums[last_inc_index] < nums[last_inc_index - 1]) {
            --last_inc_index;
        }

        if (last_inc_index == 0) {
            reverse(nums.begin(), nums.end());
            return nums;
        } else {
            int target = nums[last_inc_index - 1];
            reverse(nums.begin() + last_inc_index, nums.end());

            int min_valid_index =
                upper_bound(nums.begin() + last_inc_index, nums.end(), target) - nums.begin();
            swap(nums[min_valid_index], nums[last_inc_index - 1]);

            return nums;
        }
    }
};
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