使用一个快指针fast来遍历数组, 一个慢指针slow来记录最终结果, 用count记录同一个数出现的次数, 如果遇到新数或count<=2则更新A[slow]。

Use a "fast" pointer to traverse the array, a "slow" pointer to record final results. Use "count" to record the occurrences of a number. If "fast" meets a new number or count <= 2, then update A[slow].

class Solution:
    # @param A a list of integers
    # @return an integer
    def removeDuplicates(self, A):
        lenA = len(A)
        if lenA == 0: return 0
        slow = 0
        count = 1
        for fast in xrange(1, lenA):
            if A[fast] == A[fast - 1]:
                count += 1
                if count <= 2:
                    slow += 1
                    A[slow] = A[fast]
            else:
                count = 1
                slow += 1
                A[slow] = A[fast]
        return slow + 1