Two Pointers

算法思路：

left is left pointer, i is right pointer

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)
			<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)
	return <result>

算法分析：

时间复杂度为O(n)，空间复杂度O(1)。