1734. arrangement after decoding XOR

Give you an array of integers perm , It's before n An arrangement of positive integers , And n It's a Odd number .

It's encrypted to another length of n - 1 Array of integers for encoded , Satisfy encoded[i] = perm[i] XOR perm[i + 1] . For example , If perm = [1,3,2] , that encoded = [2,1] .

Here you are. encoded Array , Please return to the original array perm . The question guarantees that the answer exists and is unique .

Example 1：

 Input ：encoded = [3,1]
 Output ：[1,2,3]
 explain ： If  perm = [1,2,3] , that  encoded = [1 XOR 2,2 XOR 3] = [3,1]

Example 2：

 Input ：encoded = [6,5,4,6]
 Output ：[2,4,1,5,3]

Tips ：

3 <= n < 105
n  Is odd .
encoded.length == n - 1

class Solution {
    
public:
    vector<int> decode(vector<int>& encoded) {
    
        vector<int>v;
        //n Is odd 
        //len=n-1;
        int len=encoded.size();
        int n=len+1;
        
        int first=0;
        for(int i=1;i<=n;i++){
    
            first^=i;
        }
        for(int i=1;i<len;i=i+2){
    
            first^=encoded[i];
        }
        
        v.push_back(first);
        for(int i=1;i<n;i++){
    
            v.push_back(v[i-1]^encoded[i-1]);
        }
        return v;
        

    }
};