Permutations

Description

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

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

Constraints:

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

Code

backtracking

經典 backtracking 題目，使用 hashmap 是為了降低 time complexity。若沒有使用 hashmap，則對於 [1, 2, 3] 而言，結果會是：

[[1,1,1],[1,1,2],[1,1,3],[1,2,1],[1,2,2],[1,2,3],[1,3,1],[1,3,2],[1,3,3],[2,1,1],[2,1,2],[2,1,3],[2,2,1],[2,2,2],[2,2,3],[2,3,1],[2,3,2],[2,3,3],[3,1,1],[3,1,2],[3,1,3],[3,2,1],[3,2,2],[3,2,3],[3,3,1],[3,3,2],[3,3,3]]

會有重複使用的情況發生。

class Solution {
public:
    vector<vector<int>> permute(vector<int>& nums) {
        vector<vector<int>> answer;
        int n = nums.size();
        vector<int> curr;
        set<int> h;
        helper(answer, nums, curr, h);
        return answer;
    }
 
    void helper(vector<vector<int>> &answer, vector<int> nums, vector<int> curr, set<int> hashmap) {
        if(curr.size() == nums.size()) {
            answer.push_back(curr);
            return;
        }
 
        for(int i = 0; i < nums.size(); i++) {
            if(hashmap.count(nums[i]) > 0) continue;
            curr.push_back(nums[i]);
            hashmap.insert(nums[i]);
            // backtracking
            helper(answer, nums, curr, hashmap);
            curr.pop_back();
            hashmap.erase(nums[i]);
        }
        
    }
};

next permutation

由 Next Permutation 我們知道如何求出 next permutation，因此也可以直接做 n! 次 next permutation，回傳所有的結果。

class Solution {
public:
    vector<vector<int>> permute(vector<int>& nums) {
        vector<vector<int>> answer;
        int n = nums.size();
        int factorial = 1;
        for(int i = 1; i <= n; i++) {
            factorial *= i;
        }
 
        for(int i = 0; i < factorial; i++) {
            answer.push_back(nums);
            nextPermutation(nums);
        }
        return answer;
    }
 
    void nextPermutation(vector<int>& nums) {
        int n = nums.size(), k, l;
        for(k = n - 2; k >= 0; k--) {
            if(nums[k] < nums[k+1]) 
                break;
        }
 
        if(k < 0) return reverse(nums.begin(), nums.end());
 
        for(l = n - 1; l >= 0; l--) {
            if(nums[l] > nums[k]) 
                break;
        }
 
        swap(nums[l], nums[k]);
        reverse(nums.begin() + k + 1, nums.end());
    }
};

Source

• Permutations - LeetCode