Two Pointers

算法思路：

left is left pointer, i is right pointer
外循环为扩张，内循环为收缩。收缩条件为固定字符种数(必须固定或者少于k)或者(以及)固定字符频数。若字符种数不固定，要试1-26种字符，才能单调，详见L395

求最短串

Python代码：

def two_pointers(self, nums):
	for i in range(len(nums)):
		<calculate condition such as char_to_count>
		while <meets condition>:
			res = min(res, i - left + 1) <求最短子列>
			<anti-calculate condition such as char_to_count>
			left += 1
	return <result>

求最长串

Python代码：

def two_pointers(self, nums):
	for i in range(len(nums)):
		<calculate condition such as char_to_count>
		while <does not meet condition>:
			<anti-calculate condition such as char_to_count>
			left += 1
		res = max(res, i - left + 1) <求最短子列>
	return <result>

算法分析：

时间复杂度为O(n)，空间复杂度O(1)。