27. Remove Element(easy)

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.

 注意雖然需要的返回值是int，但是同時會測評你的nums是否去掉了所有val，且是inplace修改，即引用地址沒變

還有一點是val值可能不存在nums中，需要提前判斷

1 我的最優解

思路：因為題目一直提醒說剩下的nums可以不必按順序，所以機智如我，先排序一波，此時val的位置index和數量count是已知的

利用切片把從[index ：index+count]的val 都賦值為空[]

class Solution:
    def removeElement(self, nums, val):
        """
        :type nums: List[int]
        :type val: int
        :rtype: int
        """

        if not val in nums: return len(nums)
        num = len(nums) - nums.count(val)
        nums.sort()
        nums[nums.index(val):nums.index(val)+nums.count(val)] = []
        return num
 

Runtime: 36 ms, faster than 99.77% of Python3  

 2 普通思路

計算有幾個val，就刪幾次val

class Solution:
    def removeElement(self, nums, val):
        """
        :type nums: List[int]
        :type val: int
        :rtype: int
        """
        if not val in nums: return len(nums)
        num = len(nums) - nums.count(val)
        length = nums.count(val)
        for i in range(length):
            nums.remove(val)
        
        return num