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