[pull] master from wisdompeak:master #345

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		[2537.Count-the-Number-of-Good-Subarrays](https://github.com/wisdompeak/LeetCode/tree/master/Two_Pointers/2537.Count-the-Number-of-Good-Subarrays) (M+)
		[3298.Count-Substrings-That-Can-Be-Rearranged-to-Contain-a-String-II](https://github.com/wisdompeak/LeetCode/tree/master/Two_Pointers/3298.Count-Substrings-That-Can-Be-Rearranged-to-Contain-a-String-II) (M+)
		[3306.Count-of-Substrings-Containing-Every-Vowel-and-K-Consonants-II](https://github.com/wisdompeak/LeetCode/tree/master/Two_Pointers/3306.Count-of-Substrings-Containing-Every-Vowel-and-K-Consonants-II) (H-)
	[3641.Longest-Semi-Repeating-Subarray](https://github.com/wisdompeak/LeetCode/tree/master/Two_Pointers/3641.Longest-Semi-Repeating-Subarray) (H-)
		* ``Two pointers for two sequences``
		[986.Interval-List-Intersections](https://github.com/wisdompeak/LeetCode/tree/master/Two_Pointers/986.Interval-List-Intersections) (M)
		[1229.Meeting-Scheduler](https://github.com/wisdompeak/LeetCode/tree/master/Two_Pointers/1229.Meeting-Scheduler) (M+)
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	class Solution {
	public:
	int longestSubarray(vector<int>& nums, int k) {
	unordered_map<int,int>Map;
	int count = 0;
	int n = nums.size();
	int j = -1;
	int ret = 0;
	for (int i=0; i<n; i++) {
	while (j+1<n && count<=k) {
	Map[nums[++j]]++;
	if (Map[nums[j]]==2)
	count++;
	}
	if (count>k)
	ret = max(ret, j - i);
	else
	ret = max(ret, j - i + 1);
	Map[nums[i]]--;
	if (Map[nums[i]]==1) count--;
	}
	return ret;

	}
	};

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	### 3641.Longest-Semi-Repeating-Subarray

	非常明显的双指针,但是滑窗右边界j的控制需要格外小心。

	对于任意的左边界i,当满足`while (j+1<n && count<=k)`时(count变量用于统计repeat元素的个数),说明可以将nums[++j]加入统计。如果循环退出,注意有两种不同情况:
	1. 如果count>k,那么意味着j超出了合法的范围,合法subarray的长度最多是`j-i`.
	2. 如果count<=k,那么意味着j已经到了最后一个元素(即n-1)但是仍未超出合法范围,故此时合法subarray的长度是`j-i+1`.

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