80. 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.
思路：因为是排好序的数组，所以统计每个数字出现的次数，下标idx表示重新给nums数组赋值的下标，超过两次的赋值两次。小于两次的就直接赋值，更新idx和now。注意全部相同的情况（1，1，1，1），扫完数组才执行Count

class Solution {public:    void Count(vector<int>& nums, int cnt, int& idx, int now){        if(cnt >= 2){            nums[idx] = now;            nums[idx + 1] = now;            idx += 2;        } else if(cnt == 1){            nums[idx] = now;            ++idx;        }    }    int removeDuplicates(vector<int>& nums) {        int sz = nums.size();        int cnt = 0, idx = 0, now;        for(int i = 0; i < sz; ++i){            if(i == 0) now = nums[i];            if(nums[i] == now){                ++cnt;            } else {                Count(nums, cnt, idx, now);                cnt = 1;                now = nums[i];            }        }        Count(nums, cnt, idx, now);        return idx;    }};