Sponsored
Sponsored
This approach uses two pointers: one to track the position for the next non-zero element and the other to iterate through the array. We move all non-zero elements to the beginning of the array using these two pointers and fill the remaining positions with zeroes.
Time Complexity: O(n), where n is the length of the array. We make a single pass through the array.
Space Complexity: O(1), as we perform the operation in place.
1public class Solution {
2 public void moveZeroes(int[] nums) {
3 int lastNonZeroFoundAt = 0;
4 for (int i = 0; i < nums.length; i++) {
5 if (nums[i] != 0) {
6 nums[lastNonZeroFoundAt++] = nums[i];
7 }
8 }
9 for (int i = lastNonZeroFoundAt; i < nums.length; i++) {
10 nums[i] = 0;
11 }
12 }
13}In this Java solution, we use a similar approach, using an integer lastNonZeroFoundAt to track where to place the next non-zero number found as we iterate through nums.
This method uses a two-pointer technique where we place one pointer at the beginning of the array and the other to iterate through the array. Whenever we encounter a non-zero element, we swap it with the first pointer's position, allowing us to effectively move zeroes to the end by swapping.
Time Complexity: O(n), single iteration with swaps.
Space Complexity: O(1), in-place swaps.
1#include <vector>
using namespace std;
void moveZeroes(vector<int>& nums) {
int j = 0;
for (int i = 0; i < nums.size(); i++) {
if (nums[i] != 0) {
swap(nums[j++], nums[i]);
}
}
}In C++, we use the swap function to interchange the values at non-zero index i with the value at j, incrementing j each time a swap is made.