LeetCode - Permutations

最新推荐文章于 2025-12-28 17:52:30 发布

原创最新推荐文章于 2025-12-28 17:52:30 发布 · 578 阅读

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文章标签：

#leetcode #recursion #dfs

本文介绍了如何通过递归和动态规划的方法，利用一个向量来存储哪些数字已经被访问过，从而高效地生成一组数字的所有可能排列组合。详细解释了算法的实现过程，并提供了一个具体的C++代码示例，该代码可在LeetCode平台上的相关问题中应用。

Given a collection of 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].

http://oj.leetcode.com/problems/permutations/

Solution:

Recursion and dp, use a vector to store which digit has been visited.

https://github.com/starcroce/leetcode/blob/master/permuntations.cpp

// 68 ms for 25 test cases
// use a visited vector to store which digit has been visited
class Solution {
public:
    vector<vector<int> > permute(vector<int> &num) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        vector<vector<int> > res;
        if(num.size() == 0) {
            return res;
        }
        vector<int> sol;
        vector<bool> visited(num.size(), false);
        permute(num, 0, visited, res, sol);
        return res;
    }
    
    void permute(vector<int> &num, int level, vector<bool> visited, vector<vector<int> > &res, vector<int> sol) {
        if(sol.size() == num.size()) {
            res.push_back(sol);
            return;
        }
        for(int i = 0; i < num.size(); i++) {
            if(visited[i] == false) {
                visited[i] = true;
                sol.push_back(num[i]);
                permute(num, i+1, visited, res, sol);
                sol.pop_back();
                visited[i] = false;
            }
        }
    }
};