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.
Clarification:
Confused why the returned value is an integer but your answer is an array?
Note that the input array is passed in by reference, which means a modification to the input array will be known to the caller as well.
Internally you can think of this:
// nums is passed in by reference. (i.e., without making a copy) int len = removeElement(nums, val); // any modification to nums in your function would be known by the caller. // using the length returned by your function, it prints the first len elements. for (int i = 0; i < len; i++) { print(nums[i]); }
Example 1:
Input: nums = [3,2,2,3], val = 3 Output: 2, nums = [2,2] Explanation: 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. For example if you return 2 with nums = [2,2,3,3] or nums = [2,2,0,0], your answer will be accepted.
Example 2:
Input: nums = [0,1,2,2,3,0,4,2], val = 2 Output: 5, nums = [0,1,4,0,3] Explanation: Your function should return length =5
, with the first five elements ofnums
containing0
,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.
Constraints:
0 <= nums.length <= 100
0 <= nums[i] <= 50
0 <= val <= 100