Move Zeroes Solutions in C++

// Source : https://leetcode.com/problems/move-zeroes/
// Author : Calinescu Valentin, Hao Chen
// Date   : 2015-10-21





class Solution {
public:
    /* 
     * Solution (Calinescu Valentin)
     * ==============================
     *
     * One solution would be to store the position of the next non-zero element of the array.
     * Every position of the array, starting with position 0, must store this next non-zero
     * element until we can no more do that, case in which we need to add the remaining zeros
     * that we skipped.
     * 
     * 
     * Time Complexity: O(N)
     * Space Complexity: O(1)
     * 
     */
    void moveZeroes(vector<int>& nums) {
        int i = 0, poz = 0;
        for(i = 0; i < nums.size() && poz < nums.size(); i++)
        {
                while(poz < nums.size() && nums[poz] == 0)
                    poz++;
                if(poz < nums.size())
                    nums[i] = nums[poz];
                else
                    i--; // we need 0 on position i, but i is increasing one last time
                poz++;
        }
        for(; i < nums.size(); i++)
            nums[i] = 0;
    }



    /*
     * Another implemtation which is easy to understand (Hao Chen)
     * ===========================================================
     *
     * We have two pointers to travel the array, assume they named `p1` and `p2`.
     * 
     *   1) `p1` points the tail of current arrays without any ZEROs.
     *   2) `p2` points the head of the rest array which skips the ZEROs.
     * 
     * Then we can just simply move the item from `p2` to `p1`.
     *
     */
    void moveZeroes(vector<int>& nums) {
        int p1=0, p2=0;

        // Find the first ZERO, where is the tail of the array.
        // (Notes: we can simply remove this!)
        for (; p1<nums.size() && nums[p1]!=0; p1++);
        
        // copy the item from p2 to p1, and skip the ZERO
        for (p2=p1; p2<nums.size(); p2++) {
            if ( nums[p2] == 0 ) continue;
            nums[p1++] = nums[p2]; 
        }    
        
        //set ZERO for rest items 
        while ( p1<nums.size() ) nums[p1++] = 0;
    }

};

/// Source : https://leetcode.com/problems/move-zeroes/description/
/// Author : liuyubobobo
/// Time   : 2017-02-09

#include <iostream>
#include <vector>

using namespace std;

// Time Complexity: O(n)
// Space Complexity: O(n)
class Solution {
public:
    void moveZeroes(vector<int>& nums) {

        vector<int> nonZeroElements;

        // put all the non zero elements into a new vector
        for(int i = 0 ; i < nums.size() ; i ++)
            if(nums[i])
                nonZeroElements.push_back(nums[i]);

        // make nums[0...nonZeroElements.size()) all non zero elements
        for(int i = 0 ; i < nonZeroElements.size() ; i ++)
            nums[i] = nonZeroElements[i];

        // make nums[nonZeroElements.size()...nums.size()) all zero elements
        for(int i = nonZeroElements.size() ; i < nums.size() ; i ++)
            nums[i] = 0;
    }
};


void printVec(const vector<int>& vec){
    for(int e: vec)
        cout << e << " ";
    cout << endl;
}

int main() {

    int arr[] = {0, 1, 0, 3, 12};
    vector<int> vec(arr, arr + sizeof(arr) / sizeof(int));
    Solution().moveZeroes(vec);
    printVec(vec);

    return 0;
}

/// Source : https://leetcode.com/problems/move-zeroes/description/
/// Author : liuyubobobo
/// Time   : 2017-02-09

#include <iostream>
#include <vector>

using namespace std;

// Time Complexity: O(n)
// Space Complexity: O(1)
class Solution {
public:
    void moveZeroes(vector<int>& nums) {

        int k = 0; // keep nums[0...k) are all zero elements
        for(int i = 0 ; i < nums.size() ; i ++ )
            if(nums[i])
                nums[k++] = nums[i];

        // make the nums[k...end) zeros
        for(int i = k ; i < nums.size() ; i ++)
            nums[i] = 0;
    }
};


void printVec(const vector<int>& vec){
    for(int e: vec)
        cout << e << " ";
    cout << endl;
}

int main() {

    int arr[] = {0, 1, 0, 3, 12};
    vector<int> vec(arr, arr + sizeof(arr) / sizeof(int));
    Solution().moveZeroes(vec);
    printVec(vec);

    return 0;
}

/// Source : https://leetcode.com/problems/move-zeroes/description/
/// Author : liuyubobobo
/// Time   : 2017-02-09

#include <iostream>
#include <vector>

using namespace std;

// Time Complexity: O(n)
// Space Complexity: O(1)
class Solution {
public:
    void moveZeroes(vector<int>& nums) {

        int k = 0; // keep nums[0...k) are all zero elements
        for(int i = 0 ; i < nums.size() ; i ++)
            if(nums[i])
                swap(nums[k++] , nums[i]);
    }
};


void printVec(const vector<int>& vec){
    for(int e: vec)
        cout << e << " ";
    cout << endl;
}

int main() {

    int arr[] = {0, 1, 0, 3, 12};
    vector<int> vec(arr, arr + sizeof(arr) / sizeof(int));
    Solution().moveZeroes(vec);
    printVec(vec);

    return 0;
}

/// Source : https://leetcode.com/problems/move-zeroes/description/
/// Author : liuyubobobo
/// Time   : 2017-02-09

#include <iostream>
#include <vector>

using namespace std;

// Time Complexity: O(n)
// Space Complexity: O(1)
class Solution {
public:
    void moveZeroes(vector<int>& nums) {

        int k = 0; // keep nums[0...k) are all zero elements
        for(int i = 0 ; i < nums.size() ; i ++ )
            if(nums[i])
                // avoid self swapping
                if(k != i)
                    swap(nums[k++], nums[i]);
                else
                    k ++;
    }
};


void printVec(const vector<int>& vec){
    for(int e: vec)
        cout << e << " ";
    cout << endl;
}

int main() {

    int arr[] = {0, 1, 0, 3, 12};
    vector<int> vec(arr, arr + sizeof(arr) / sizeof(int));
    Solution().moveZeroes(vec);
    printVec(vec);

    return 0;
}