problem:

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].

这个题目仅仅考察数组操作，没有难点。

solution：

法一

class Solution:
    # @param A a list of integers
    # @return an integer
    def removeDuplicates(self,A):
        n = len(A)
        if n ==0 :
            return 0
        index = 0
        times = 0
        for i in xrange(n):
            if A[index] == A[i]:
                times += 1
                if times ==2 :
                    index += 1
                    A[index] = A[i]
            else:
                index += 1
                A[index] = A[i]
                times = 1
        A = A[:index+1]
        return len(A)
法二 
class Solution:
    # @param A a list of integers
    # @return an integer
    def removeDuplicates(self,A):
        n = len(A)
        if n <= 2 :
            return n
        index = 2
    
        for i in xrange(2,n):
            if A[i] != A[index-2]:
                A[index] = A[i]
                index += 1
            
        A = A[:index]
        return len(A)