Given an integer array nums, move all 0's to the end of it while maintaining the relative order of the non-zero elements.
Note that you must do this in-place without making a copy of the array.
Example 1:
Input: nums = [0,1,0,3,12]
Output: [1,3,12,0,0]
Example 2:
Input: nums = [0]
Output: [0]
Constraints:
1 <= nums.length <= 104
-231 <= nums[i] <= 231 - 1
Follow up: Could you minimize the total number of operations done?
Solutions
Solution 1: Two Pointers
We use a pointer $k$ to record the current position to insert, initially $k = 0$.
Then we iterate through the array $\textit{nums}$, and each time we encounter a non-zero number, we swap it with $\textit{nums}[k]$ and increment $k$ by 1.
This way, we can ensure that the first $k$ elements of $\textit{nums}$ are non-zero, and their relative order is the same as in the original array.
The time complexity is $O(n)$, where $n$ is the length of the array $\textit{nums}$. The space complexity is $O(1)$.
/** Do not return anything, modify nums in-place instead. */functionmoveZeroes(nums:number[]):void{letk=0;for(leti=0;i<nums.length;++i){if(nums[i]){[nums[i],nums[k]]=[nums[k],nums[i]];++k;}}}