原题地址：https://oj.leetcode.com/problems/remove-duplicates-from-sorted-array-ii/
题意：
Follow up for "Remove Duplicates":
What if duplicates are allowed at most twice?
For example,
Given sorted array A = [1,1,1,2,2,3],
Your function should return length = 5, and A is now [1,1,2,2,3].
解题思路：一种巧妙的解法。使用两个指针prev和curr，判断A[curr]是否和A[prev]、A[prev-1]相等，如果相等curr指针继续向后遍历，直到不相等时，将curr指针指向的值赋值给A[prev+1]，这样多余的数就都被交换到后面去了。最后prev+1值就是数组的长度。
代码：
class Solution:
# @param A a list of integers
# @return an integer
# @it's a good solution!
def removeDuplicates(self, A):
if len(A) <= 2: return len(A)
prev = 1; curr = 2
while curr < len(A):
if A[curr] == A[prev] and A[curr] == A[prev - 1]:
curr += 1
else:
prev += 1
A[prev] = A[curr]
curr += 1
return prev + 1