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 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:
Given nums = [3,2,2,3], val = 3,
Your function should return length = 2, with the first two elements of nums being 2.
class Solution:
def removeElement(self, nums, val):
"""
:type nums: List[int]
:type val: int
:rtype: int
"""
i = 0
for j in range(len(nums)):
if nums[j] != val:
nums[i] = nums[j]
i += 1
return i
Complexity analysis:
Time complexity : O(n). Assume the array has a total of n elements, both i and j traverse at most 2n steps.
Space complexity : O(1).
