permutation and its variations

最新推荐文章于 2023-08-10 09:22:39 发布

原创最新推荐文章于 2023-08-10 09:22:39 发布 · 319 阅读

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本文深入探讨了两种排列算法：一种用于生成所有可能的排列，另一种用于生成唯一排列，避免重复。通过递归深度优先搜索（DFS）实现，文章详细介绍了算法的具体实现过程，并展示了如何处理重复元素。

Permutations

If we change the argument $vector<int> nums$

class Solution {
public:
    vector<vector<int>> permute(vector<int>& nums) {
        vector<vector<int>> res;
        if(nums.empty())    return res;
        sort(nums.begin(), nums.end());
        DFS(res, nums, 0);
        return res;
    }

    void DFS(vector<vector<int>>& res, vector<int> nums, int start){
        if(start == nums.size() - 1){
            res.push_back(nums);
            return;
        }
        for(int i = start; i < nums.size(); i++){
            swap(nums[i], nums[start]);
            DFS(res, nums, start + 1);
        }
    }
};

If we change the argument $vector<int> nums$ into reference, we would need another swap:

swap(nums[i], nums[start]);
DFS(res, nums, start + 1);
swap(nums[i], nums[start]);

Permutations II

class Solution {
public:
    vector<vector<int>> permuteUnique(vector<int>& nums) {
        vector<vector<int>> res;
        if(nums.empty())    return res;
        sort(nums.begin(), nums.end());
        DFS(res, nums, 0);
        return res;
    }

    void DFS(vector<vector<int>>& res, vector<int> nums, int start){
        if(start == nums.size() - 1){
            res.push_back(nums);
            return;
        }
        for(int i = start; i < nums.size(); i++){
            if(i != start && nums[i] == nums[start])
                continue;
            swap(nums[i], nums[start]);
            DFS(res, nums, start + 1);
        }
    }
};

If we change the argument $vector<int> nums$ into reference, we would need a few lines of modifications:

    void helper(vector<vector<int>>& res, vector<int>& nums, int pos) {

        if (pos == nums.size()) {
            res.push_back(nums);
        } else {
            for (int i = pos; i < nums.size(); ++i) {
                if (i > pos && nums[i] == nums[pos]) continue;
                swap(nums[pos], nums[i]);
                helper(res, nums, pos + 1);

            }
            // restore nums
            for (int i = nums.size() - 1; i > pos; --i) {
                swap(nums[pos], nums[i]);
            }
        }
    }