LeetCode 47. Full arrangement II

Title Description

47. Full Permutation II

solution

This problem is also the problem of backtracking pruning , But there are some skills in pruning here . Let's analyze first , for example ,$nums=[1,2,2^{\prime }]$, Without pruning , We will get the following results

[
    [1,2,2'],[1,2',2],
    [2,1,2'],[2,2',1],
    [2',1,2],[2',2,1]
]

If we ensure that the relative positions of the same elements in the arrangement remain unchanged , for instance $nums=[1,2,2^{\prime }]$ This example , I keep in line $2$ Has been $2^{\prime }$ front . In this case , You from above 6 You can only pick out 3 Two permutations meet this condition ：

[ 
	[1,2,2'],[2,1,2'],[2,2',1] 
]

Think about , It should be easy to understand the principle ：

The reason why the standard full permutation algorithm repeats , Because the arrangement sequence formed by the same elements is regarded as different sequences , But actually they should be the same ; And if you fix the sequence order formed by the same elements , Of course, it avoids repetition

Then reflect it on the code , Pay attention to the pruning logic ：

//  Newly added pruning logic , Fix the relative position of the same element in the arrangement 
if (i > 0 && nums[i] == nums[i - 1] && !used[i - 1]) {
    
    //  If the previous adjacent equal elements have not been used , Then skip 
    continue;
}
//  choice  nums[i]

The complete implementation is as follows

class Solution {
    
public:

    vector<vector<int>> res;

    vector<vector<int>> permuteUnique(vector<int>& nums) {
    
        vector<int> track;
        vector<int> used(nums.size(), 0);
        sort(nums.begin(), nums.end());
        backtrace(nums, track, used);
        return res;
    }

    void backtrace(vector<int>& nums, vector<int>& track, vector<int>& used)
    {
    
        if (track.size() == nums.size())
        {
    
            res.push_back(track);
            return;
        }

        for (int i = 0; i < nums.size(); i++)
        {
    
            if (used[i]) continue;
            if (i > 0 && nums[i] == nums[i - 1] && used[i - 1] == 0) continue;

            track.push_back(nums[i]);
            used[i] = 1;
            backtrace(nums, track, used);
            used[i] = 0;
            track.pop_back();
        }
    }
};