80. Remove Duplicates from Sorted Array II

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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.
思路：这道题目是 26.Remove Duplicates from Sorted Array 的follow-up。不同之处在于，可以允许每个元素重复最多两次。所以，我们还是继续使用i和index这两个下标，只不过不同之处在于，i和index不再从0开始，而是都从2开始。然后每次判断是否满足nums[i] != nums[index - 2]，如果满足，就把nums[i]的值赋给nums[index]。

需要注意：
1.如果nums.size() <= 2，就直接返回nums.size()；
2.最后返回值为index即可。

代码如下：

class Solution {public:    int removeDuplicates(vector<int>& nums) {        if(nums.size() <= 2) return nums.size();        int index = 2;        for(int i = 2; i < nums.size(); ++i){            if(nums[i] != nums[index - 2])                nums[index++] = nums[i];        }        return index;    }};