LeetCode 46. Permutations

最新推荐文章于 2023-12-02 16:16:52 发布

原创最新推荐文章于 2023-12-02 16:16:52 发布 · 292 阅读

0 ·

CC 4.0 BY-SA版权

LeetCode 题解同时被 2 个专栏收录

288 篇文章

订阅专栏

Backtrack

5 篇文章

订阅专栏

本文介绍了一个经典的排列生成算法实现，该算法通过递归回溯的方式，对给定的一组不重复数字生成所有可能的排列组合，并确保输出结果有序。文章提供了详细的C++代码示例，演示如何使用标记数组避免元素重复使用。

Given a collection of distinct numbers, return all possible permutations.

For example,
[1,2,3] have the following permutations:
[1,2,3], [1,3,2], [2,1,3], [2,3,1], [3,1,2], and [3,2,1].

Classical backtrack.

In order to make the output in sequence, sort the inputs first.

   void permute(vector<int>& nums, int depth, vector<vector<int>>& res, vector<int>& path, vector<int>& used) {
        if(depth == nums.size()) res.push_back(path);  // depth variable can be saved, if here checks path.size() == nums.size()
        for(int i = 0; i < nums.size(); ++i) {
            if(used[i] == false) {
                used[i] = true;                        // use a flag to mark that the value has been used.
                path.push_back(nums[i]);
                permute(nums, depth + 1, res, path, used);
                used[i] = false;
                path.pop_back();
            }
        }
    }
    vector<vector<int>> permute(vector<int>& nums) {
        if(nums.size() == 0) return {};
        sort(nums.begin(), nums.end());
        vector<vector<int>> res;
        vector<int> path;
        vector<int> used(nums.size(), false);
        permute(nums, 0, res, path, used);
        return res;
    }