27. Remove Element

problem

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.

Example:
Given input array nums = [3,2,2,3], val = 3

Your function should return length = 2, with the first two elements of nums being 2.



Hint:

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?

删除列表中给定的元素，返回删除后的列表长度n

• 列表中元素要删除，或者更改序列：列表前n个元素为理论上删除后的列表

solution

列表删除

class Solution(object):
    def removeElement(self, nums, val):
        """
        :type nums: List[int]
        :type val: int
        :rtype: int
        """
        length = len(nums)
        for i in range(length-1,-1,-1):
            if nums[i] == val:
                del nums[i]
        
        return len(nums)

改变列表排序

class Solution(object):
    def removeElement(self, nums, val):
        """
        :type nums: List[int]
        :type val: int
        :rtype: int
        """
        index = 0
        length = len(nums)
        for i in range(length):
            if nums[i] != val:
                nums[index] = nums[i] 
                index += 1
                
        return index

如果不限行空间，还可考虑用Counter（总数量-val数量） 不行，列表本身要变化