題目：

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.

Example 1:

Given nums = [3,2,2,3], val = 3,

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.

Example 2:

Given nums = [0,1,2,2,3,0,4,2], val = 2,

Your function should return length = 5, with the first five elements of nums

Note that the order of those five elements can be arbitrary.

It doesn't matter what values are set beyond the returned 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

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]); }

解答：

由於這個nums是list[int]型別，而不是python核心資料型別，直接for迴圈刪除是不行的。

要從最後一個數據開始處理，否則不能pass

class Solution:
    def removeElement(self, nums, val):
        """
        :type nums: List[int]
        :type val: int
        :rtype: int
        """
        length = len(nums)
        while length>0:
            if nums[length-1]==val:
                nums.pop(length-1)
            length -= 1
        
        return len(nums)