标签：mat fast rem 进入 input contain hat Plan span
Given a string, find the length of the longest substring without repeating characters.
Example 1:
Input: "abcabcbb"
Output: 3
Explanation: The answer is "abc", with the length of 3.
Example 2:
Input: "bbbbb"
Output: 1
Explanation: The answer is "b", with the length of 1.
Example 3:
Input: "pwwkew"
Output: 3
Explanation: The answer is "wke", with the length of 3.
Note that the answer must be a substring, "pwke" is a subsequence and not a substring.
M1: sliding window + hash map (more generalize)
用d表示所求substring长度。fast扫描string，存进map。用counter计数，如果map中对应value > 1，counter++。当counter > 0时进入while循环开始移动slow找有效的substring，如果slow所指的字符对应出现次数 > 1，counter--，并存入新的value，slow向右移动。退出while循环时即找到有效substring，此时记录fast - slow，并和d比较，取较大值。
time: O(n), space: O(n)
class Solution { public int lengthOfLongestSubstring(String s) { Map<Character, Integer> map = new HashMap<>(); int slow = 0, fast = 0, counter = 0, d = 0; while(fast < s.length()) { char c = s.charAt(fast); map.put(c, map.getOrDefault(c, 0) + 1); if(map.get(c) > 1) { counter++; } fast++; while (counter > 0) { char tmp = s.charAt(slow); if (map.get(tmp) > 1) { counter--; } map.put(tmp, map.get(tmp) - 1); slow++; } d = Math.max(d, fast - slow); } return d; } }
M2: sliding window + hash set (specific to unique character)
time: O(n), space: O(n)
class Solution { public int lengthOfLongestSubstring(String s) { Set<Character> set = new HashSet<>(); int slow = 0, fast = 0, max = 0; while(fast < s.length()) { if(!set.contains(s.charAt(fast))) { max = Math.max(max, fast - slow + 1); set.add(s.charAt(fast++)); } else { set.remove(s.charAt(slow++)); } } return max; } }
3. Longest Substring Without Repeating Characters - Medium
标签：mat fast rem 进入 input contain hat Plan span
原文地址：https://www.cnblogs.com/fatttcat/p/10301890.html
