[LintCode] Permutations

http://www.lintcode.com/en/problem/permutations/#

Given a list of numbers, return all possible permutations.

Example

For nums = [1,2,3], the permutations are:

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

求全排列,可以使用DFS来解决,来看代码:

class Solution {
public:
    /**
     * @param nums: A list of integers.
     * @return: A list of permutations.
     */
    vector<vector<int> > permute(vector<int> nums) {
        // write your code here
        vector<vector<int>> paths;
        if (nums.empty()) {
            return paths;
        }
        
        vector<int> index;
        vector<int> path;
        permuteHelper(nums, index, path, paths);
        return paths;
        
    }
    
private:
    void permuteHelper(const vector<int> &nums,
                       vector<int> &index,    
                       vector<int> &path,
                       vector<vector<int>> &paths) {
        if (path.size() == nums.size()) {
            paths.push_back(path);
            return;
        }
        
        for (int ix = 0; ix < nums.size(); ix++) {
            if (find(index.begin(), index.end(), ix) == index.end()) {
                index.push_back(ix);
                path.push_back(nums[ix]);
                permuteHelper(nums, index, path, paths);
                index.pop_back();
                path.pop_back();
            }
        }                   
    }
};

实际上,观察某数是否已经访问过,不必使用一个vector,因为在vector中看一个数有没有访问过,需要o(n)的时间复杂度,此处完全可以用一个hashset来代替,看以下代码:

class Solution {
public:
    /**
     * @param nums: A list of integers.
     * @return: A list of permutations.
     */
    vector<vector<int> > permute(vector<int> nums) {
        // write your code here
        vector<vector<int>> paths;
        if (nums.empty()) {
            return paths;
        }
        
        unordered_set<int> index;
        vector<int> path;
        permuteHelper(nums, index, path, paths);
        return paths;
    }
    
private:
    void permuteHelper(const vector<int> &nums,
                 unordered_set<int> &index,
                 vector<int> &path,
                 vector<vector<int>> &paths) {
        if (path.size() == nums.size()) {
            paths.push_back(path);
            return;
        }       
        
        for (int ix = 0; ix < nums.size(); ix++) {
            if (index.count(ix) == 0) {
                index.insert(ix);
                path.push_back(nums[ix]);
                permuteHelper(nums, index, path, paths);
                index.erase(ix);
                path.pop_back();
            }
        }
    }
};

能不能更进一步?这边完全可以使用一个数组来模拟hashset,来看代码:

class Solution {
public:
    /**
     * @param nums: A list of integers.
     * @return: A list of permutations.
     */
    vector<vector<int> > permute(vector<int> nums) {
        // write your code here
        vector<vector<int>> paths;
        if (nums.empty()) {
            return paths;
        }
        
        bool *visited = new bool[nums.size()]();
        vector<int> path;
        permuteHelper(nums, visited, path, paths);
        return paths;
    }
    
private:
    void permuteHelper(const vector<int> &nums,
                       bool *visited,
                       vector<int> &path,
                       vector<vector<int>> &paths) {
        if (path.size() == nums.size()) {
            paths.push_back(path);
            return;
        }       
        
        for (int ix = 0; ix < nums.size(); ix++) {
            if (visited[ix] == false) {
                visited[ix] = true;
                path.push_back(nums[ix]);
                permuteHelper(nums, visited, path, paths);
                visited[ix] = false;
                path.pop_back();
            }
        }
    }
};

 

posted @ 2015-05-22 22:34  Acjx  阅读(840)  评论(0编辑  收藏  举报