LeetCode-80. Remove Duplicates from Sorted Array II

Given a sorted array nums, remove the duplicates in-place such that duplicates appeared at most twice and return the new length.

Do not allocate extra space for another array, you must do this by modifying the input array in-place

Example 1:

Given 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 respectively. It doesn't matter what you leave beyond the returned length.

Example 2:

Given nums = [0,0,1,1,1,1,2,3,3], Your function should return length = ​​7​​​, with the first seven elements of​​nums​​​being modified to​​0​​, 0, 1, 1, 2, 3 and 3 respectively. It doesn't matter what values are set beyond the returned length.

Clarification:

Confused why the returned value is an integer but your answer is an array?

Note that the input array is passed in by reference, which means modification to the input array will be known to the caller as well.

Internally you can think of this:

// nums is passed in by reference. (i.e., without making a copy)
int len = removeDuplicates(nums);

// any modification to nums in your function would be known by the caller.
// using the length returned by your function, it prints the first len elements.
for (int i = 0; i < len; i++) {
    print(nums[i]);
}

题解：

两个指针遍历，第一个指针i遍历nums，第二个cnt记录不超过2个的新数组nums

因为是有序的，所以当nums[cnt - 2] == nums[i]时，nums[i]必等于nums[cnt - 1]，则此个nums[i]不放入cnt更新的nums中。

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