Two Pointers #

• Classic way to write the two-pointer sliding window. The right pointer keeps moving to the right until it can no longer move right (the specific condition depends on the problem). After the right pointer reaches the far right, start moving the left pointer to release the left boundary of the window. Problem 3, Problem 76, Problem 209, Problem 424, Problem 438, Problem 567, Problem 713, Problem 763, Problem 845, Problem 881, Problem 904, Problem 978, Problem 992, Problem 1004, Problem 1040, Problem 1052.

	left, right := 0, -1

	for left < len(s) {
		if right+1 < len(s) && freq[s[right+1]-'a'] == 0 {
			freq[s[right+1]-'a']++
			right++
		} else {
			freq[s[left]-'a']--
			left++
		}
		result = max(result, right-left+1)
	}

• Fast and slow pointers can be used to find duplicate numbers, with time complexity O(n), Problem 287.
• After replacing letters, the longest possible length of consecutive identical letters. Problem 424.
• SUM problem set. Problem 1, Problem 15, Problem 16, Problem 18, Problem 167, Problem 923, Problem 1074.