LeetCode – Remove ElementTags: Algorithm, LeetCode, python
Given an array and a value, 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 in place with constant memory.
The order of elements can be changed. It doesn’t matter what you leave beyond the new length.
Given input array nums =
[3,2,2,3], val =
Your function should return length = 2, with the first two elements of nums being 2.
- Try two pointers.
- Did you use the property of “the order of elements can be changed”?
- What happens when the elements to remove are rare?
We use two pointers: left_index and right_index. Initially, left_index points to the beginning and right_index points to the end.
If the number pointed by left_index equal to the target value, we replace it with the number pointed by right_index, and move right_index to left by one step.
Otherwise, we just move the left_index right by one step.
At the end of the program, right_index points to the last elements of the new array.
The following python code shows the implementation.
def remove_element(nums, val):
right = len(nums) - 1
left = 0
while left <= right:
if nums[left] == val:
nums[left] = nums[right]
right -= 1
left += 1
return right + 1