Overview
There are two questions I found related to this solution
Move Zeroes
Given an array nums, write a function to move all 0’s to the end of it while maintaining the relative order of the non-zero elements.
Example:
Input: [0,1,0,3,12]
Output: [1,3,12,0,0]
Note:
- You must do this in-place without making a copy of the array.
- Minimize the total number of operations.
Solution:
- Use znum to count the occurrence of zero
- if nums[i] is not 0, it should move to nums[i - znum].
- append zeros from the end of the array to nums[len - znum].
Time complexity: O(n+k): iterate over the array once and append k zeros to the array.
Space complexity: O(1): didn’t use extra array.
class Solution {
public void moveZeroes(int[] nums) {
int znum = 0, len = nums.length;
for (int i = 0; i < len; i++) {
if (nums[i] == 0) {
++znum;
} else {
nums[i - znum] = nums[i];
}
}
for (int i = len -1; i > len - 1 - znum; --i) {
nums[i] = 0;
}
}
}
Remove Element
Given an array nums and a value val, remove all instances of that value in-place and return the new length.
Do not allocate extra space for another array, you must do this by modifying the input array in-place with O(1) extra memory.
The order of elements can be changed. It doesn’t matter what you leave beyond the new length.
Example 1:
Given nums = [3,2,2,3], val = 3,
Your function should return length = 2, with the first two elements of nums being 2.
It doesn’t matter what you leave beyond the returned length.
Example 2:
Given nums = [0,1,2,2,3,0,4,2], val = 2,
Your function should return length = 5, with the first five elements of nums containing 0, 1, 3, 0, and 4.
Note that the order of those five elements can be arbitrary.
It doesn’t matter what values are set beyond the returned length.
Solution:
- Use vnum to count the occurrence of target val.
- if nums[i] is not val, it should move to nums[i - vnum].
- the new length of the array is len - vnum.
Time complexity: O(n): iterate over the array once.
Space complexity: O(1): didn’t use extra array.
class Solution {
public int removeElement(int[] nums, int val) {
int vnum = 0, len = nums.length;
for (int i = 0; i < len; i++) {
if (nums[i] == val) {
++vnum;
} else {
nums[i - vnum] = nums[i];
}
}
return len - vnum;
}
}
