80 - Remove Duplicates from Sorted Array II

原创于 2016-03-17 14:51:13 发布 · 320 阅读

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本文探讨了如何在已排序的数组中，处理最多允许出现两次重复元素的情况，通过增加计数机制来优化算法实现。具体示例包括数组[1,1,1,2,2,3]，最终返回长度为5的数组，包含1,1,2,2,3等元素。

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.

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思路分析：

增加一个count计数。

class Solution
{
public:
	int removeDuplicates(vector<int>& nums) 
	{
		int n = nums.size();

		if (n == 0)
		{
			return 0;
		}

		int key = nums[0];
		int count = 0;
		int start = 0;

		for (int i = 0; i < n; i++)
		{
			if (key == nums[i])
			{
				count++;
			}
			else
			{
				for (int j = 0; j < min(2, count); j++)
				{
					nums[start++] = key;
				}

				key = nums[i];
				count = 1;
			}
		}

		for (int j = 0; j < min(2, count); j++)
		{
			nums[start++] = key;
		}
		return start;

	}
};