Given an integer array nums
and an integer val
, remove all occurrences of val
in nums
in-place. The relative order of the elements may be changed.
Since it is impossible to change the length of the array in some languages, you must instead have the result be placed in the first part of the array nums
. More formally, if there are k
elements after removing the duplicates, then the first k
elements of nums
should hold the final result. It does not matter what you leave beyond the first k
elements.
Return k
after placing the final result in the first k
slots of nums
.
Example 1:
Input: nums = [3,2,2,3], val = 3
Output: 2, nums = [2,2,_,_]
Explanation: Your function should return k = 2, with the first two elements of nums being 2.
It does not matter what you leave beyond the returned k (hence they are underscores).
The same direction of two pointers:
Because “in-place”, I could not create a new array to keep numbers. One pointer point the position is to be covered. The other one is the index of traversal. When traversal done, end the loop.
雙指針同向
重點在in-place,所以不能開新的array來存,
那就用雙指針的方式,
一個指針指在要被覆蓋的位置,
一個指針指在要歷遍的位置,
當歷遍完成,就結束這個loop。