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.

 

 // 时间复杂度O(n)
 1 class Solution
{
public:
    int removeDuplicates(vector<int>& nums)
    {
        if (nums.size() <= 2) // 小于俩数直接返回
        {
            return nums.size();
        }

        int index = 2; // 代表已经去重的nums下标

        for (int i = 2; i < nums.size(); ++i) // 从第三个数开始遍历 因为允许两个数重复
        {
            if (nums[i] != nums[index - 2])  // 如果此时index之前两个数不与nums[i]重复就覆盖index这个单元
            {
                nums[index++] = nums[i];
            }
        }

        return index;
    }
};

 

// 时间复杂度 0(n)
class Solution
{
public:
    int removeDuplicates(vector<int>& nums)
    {
        int index = 0;
        const int n = nums.size();

        /*
            index和i均从0开始 仅当出现三个重复数时才只增加i的值
        */
        for (int i = 0; i < n; ++i)
        {
            if (i > 0 && i < n - 1 && nums[i - 1] == nums[i] && nums[i] == nums[i + 1])
            {
                continue;
            }

            nums[index++] = nums[i];
        }

        return index;
    }
};

 

// 时间复杂度 0(n)
class Solution
{
public:
    int removeDuplicates(vector<int>& nums)
    {
        int index = 0;

        for (auto n : nums)
        {
            if (index < 2 || n > nums[index - 2])
            {
                nums[index++] = n;
            }
        }

        return index;
    }
};