*leetcode #47 in cpp

最新推荐文章于 2026-01-05 22:36:07 发布

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本文介绍了一种解决排列组合中去除重复元素的方法。通过先对数组进行排序，再使用递归生成排列，过程中跳过相同的元素以避免产生重复的排列结果。此方法与问题46类似，但特别强调了如何避免生成重复的解。

The solution is very similar to the solution of $46. The only thing we should do is to avoid duplicates.

Analysis to avoid duplicates is almost the same as the one in #40.

Code:

class Solution {
public:
    vector<vector<int>> permuteUnique(vector<int>& nums) {
        sort(nums.begin(), nums.end());
        vector<vector<int>> res;
        vector<int> members;
        find_permutation(members, nums, &res);
        return res;
        
    }
    
    void find_permutation(vector<int> member, vector<int> nums, vector<vector<int>> *res){
        if(nums.size() == 1){
            member.push_back(nums[0]);
            res->push_back(member);
            return;
        }
        if(nums[0] == nums[nums.size() - 1]){//when numbers all the same, stop generating permutation
            for(int i = 0; i < nums.size(); i ++)
                member.push_back(nums[i]);
            res->push_back(member);
            return;
        }
        for(int i = 0; i < nums.size(); i ++){
            if(i > 0 && nums[i] == nums[i-1]){//avoid duplicates. skipp duplicates numbers
                continue;
            }
            int temp = nums[i];
            member.push_back(temp);
            nums.erase(nums.begin() + i);
            find_permutation(member, nums, res);
            member.pop_back();
            nums.insert(nums.begin() + i, temp);
        }
    };
    
};