L1-删除排序数组中的重复数字 II

跟进“删除重复数字”：

如果可以允许出现两次重复将如何处理？

代码

class Solution:
    """
    @param: nums: An ineger array
    @return: An integer
    """
    #def removeDuplicates(self, nums):
        # write your code here
    def removeDuplicates(self,A):
    	i = 1
    	j = 0
    	count = 1
    	while(i <len(A)):
    		if A[i] == A[j] and count == 2:
    			A.pop(i)
    		elif A[i] == A[j]:
    			count+=1
    			j+=1
    			i+=1  //注意i+=1不能放在最外面
    		else:
    			count = 1
    			j+=1
    			i+=1

或者

class Solution:  
    """ 
    @param A: a list of integers 
    @return an integer 
    """  
    def removeDuplicates(self, A):  
        if len(A) <= 1:  
            return len(A)  
        temp = A[0]  
        count, index = 1, 1  
        while index != len(A):  
            if A[index] == temp and count == 2:  
                A.pop(index)  
            elif A[index] == temp:  
                count += 1  
                index += 1  
            else:  
                temp = A[index]  
                count = 1  
                index += 1  
        return index + 1  
        # write your code here

思路

同上题，快慢指针的思想，用i来遍历数组，j来检查重复元素
如果相同且count已经为2，则移出遍历的该元素
如果相同出现一次，count++,快慢指针同时后移
如果不同，则count置1，快慢指针同时后移