版权声明：本文为博主原创文章，未经博主允许不得转载。 https://blog.csdn.net/jiang425776024/article/details/88064419
'''
Given an array and a value, remove all occurrences of that value in place and return the new length.
The order of elements can be changed, and the elements after the new length don't matter.
Example
Given an array [0,4,4,0,0,2,4,4], value=4
return 4 and front four elements of the array is [0,0,0,2]
'''
'''
一句话总结：拿两个指示变量i,j从0开始，i只负责遍历所有的值，j负责保存不包含elem的索引，i循环完了j的长度就是新的数组长度，新的数组内容就是0:j
'''
class Solution:
def call(self, arr, elem):
i = 0
j = 0
while i < len(arr):
# 如果相当则忽略，进入下一个位置
if arr[i] == elem:
pass
else: # 如果不等，则加入新的以left为长度计算的arr
arr[j] = arr[i]
j += 1 # left长度加以
i += 1 # 进入下一轮判断
return (j, arr[0:j])
s = Solution()
print(s.call([0, 4, 4, 0, 0, 2, 4, 4, 1, 2, 3, 4, 8], 4))
#(8, [0, 0, 0, 2, 1, 2, 3, 8])