有序数组去重2--Remove Duplicates from Sorted Array II
https://leetcode.com/problems/remove-duplicates-from-sorted-array-ii/
Remove Duplicates from Sorted Array II
Follow up for "Remove Duplicates":
What if duplicates are allowed at most twice?
For example,
Given sorted array nums = [1,1,1,2,2,3],
Your function should return length = 5, with the first five elements of nums being 1, 1, 2, 2 and 3. It doesn't matter what you leave beyond the new length.
题意：有序数组，去重，允许至多有2个重复项
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，如果相等说明curr指针指向的是第3个重复数，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]，最后prev+1值就是数组的长度。pre指针控制最终的数组形式。
1 class Solution:
2 # @param {integer[]} nums
3 # @return {integer}
4 def removeDuplicates(self, nums):
5 A=nums
6 if len(A)<=2: return len(A) #因允许最多2个重复,则<=2时不用去重
7 prev=1;curr=2 #用pre来控制最终的数组,curr来遍历数组进行判断
8 while curr<=len(A)-1:
9 if A[curr]==A[prev] and A[curr]==A[prev-1]: #若curr指向第3个重复数,则pre控制在第2个重复处,curr继续遍历
10 curr+=1
11 else: #若没有重复或重复在2内,将当前curr数加到pre后,pre向后移动继续控制住
12 prev+=1
13 A[prev]=A[curr]
14 curr+=1
15 return prev+1 #返回最终数组的pre+1即长度
