# -*- coding: utf8 -*-
'''
__author__ = 'dabay.wang@gmail.com'

27: Remove Element
https://oj.leetcode.com/problems/remove-element/

Given an array and a value, remove all instances of that value in place and return the new length.
The order of elements can be changed. It doesn't matter what you leave beyond the new length.

===Comments by Dabay===
一次循环,两个指针,一个指向插入的位置,另外一个一直往前面走。
如果二号指针的数就是需要删除的数,二号指针继续走。
如果不是要删除的数,把二号指针指向的数移到一号指针的位置上,然后两个指针继续走。
最后跟新数组,返回长度。
'''

class Solution:
# @param A a list of integers
# @param elem an integer, value need to be removed
# @return an integer
def removeElement(self, A, elem):
i = j = 0
while j < len(A):
if A[j] != elem:
A[i] = A[j]
i += 1
j += 1
A = A[:i]
return len(A)


def main():
sol = Solution()
print sol.removeElement([1,1,2], 1)


if __name__ == "__main__":
import time
start = time.clock()
main()
print "%s sec" % (time.clock() - start)