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LeetCode 80:Remove Duplicates from Sorted Array II

时间:2015-06-15 22:16:46      阅读:103      评论:0      收藏:0      [点我收藏+]

标签:leetcode

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.


此题要求数组中每个元素出现的个数不大于两次。所以需要加一个计数器来统计每个元素出现的次数。

代码如下:

	int removeDuplicates2(vector<int>& nums) {
		int length = nums.size();
		if (length == 0)
			return 0;
		int j=0;
		int times = 1;
		for (int i=1; i<length; i++)
		{
			if (nums[i] != nums[j])
			{
				nums[++j] = nums[i];
				times = 1;
			}else{
				times++;
				if (times == 2)
				{
					nums[++j] = nums[i];
				}
			}
		}
		return j+1;
	}



LeetCode 80:Remove Duplicates from Sorted Array II

标签:leetcode

原文地址:http://blog.csdn.net/sunao2002002/article/details/46507173

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