[LeetCode] Permutations 解题报告

递归生成所有排列算法详解

最新推荐文章于 2019-12-24 12:45:22 发布

codingtmd_in_china

最新推荐文章于 2019-12-24 12:45:22 发布

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本文链接：https://blog.youkuaiyun.com/codingtmd_in_china/article/details/50502587

本文详细介绍了如何使用递归方法生成一组数字的所有排列，并提供了C/C++代码实现。通过标志位记录已使用数字，确保生成不重复的排列组合。

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].

» Solve this problem

[解题思路]
递归，用标志位记录已使用数字。

[Code]

1:    vector<vector<int> > permute(vector<int> &num) {  
2:      // Start typing your C/C++ solution below  
3:      // DO NOT write int main() function  
4:      vector<vector<int> > coll;  
5:      vector<int> solution;  
6:      if(num.size() ==0) return coll;  
7:      vector<int> visited(num.size(), 0);  
8:      GeneratePermute(num, 0, visited, solution, coll);  
9:      return coll;  
10:    }  
11:    void GeneratePermute(vector<int> & num, int step, vector<int>& visited, vector<int>& solution, vector<vector<int> >& coll)  
12:    {  
13:      if(step == num.size())  
14:      {  
15:        coll.push_back(solution);  
16:        return;  
17:      }  
18:      for(int i =0; i< num.size(); i++)  
19:      {  
20:        if(visited[i] == 0)  
21:        {  
22:          visited[i] = 1;  
23:          solution.push_back(num[i]);  
24:          GeneratePermute(num, step+1, visited, solution, coll);  
25:          solution.pop_back();  
26:          visited[i] =0;  
27:        }  
28:      }  
29:    }