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;}}}