屏幕快照 2022-04-10 下午2.19.02.png

思路：存下每個(gè)char出現(xiàn)的次數(shù)
時(shí)間復(fù)雜度O(2n)
空間復(fù)雜度O(min(m,n))
ps：m是char的size

class Solution(object):
    def lengthOfLongestSubstring(self, s):
        """
        :type s: str
        :rtype: int
        """
        left = right = result = 0
        map = {}
        while (right < len(s)):
            map[s[right]] = 1 if s[right] not in map else map[s[right]] + 1
            while map[s[right]] > 1:
                map[s[left]] = map[s[left]] - 1
                left += 1
            result = max(result, right-left+1)
            right += 1
        return result

思路：存下每個(gè)char出現(xiàn)的index
時(shí)間復(fù)雜度O(n)
空間復(fù)雜度O(min(m,n))

class Solution(object):
    def lengthOfLongestSubstring(self, s):
        """
        :type s: str
        :rtype: int
        """
        left = right = result = 0
        map = {}
        while (right < len(s)):
            if s[right] in map and map[s[right]] >= left:
                left = map[s[right]] + 1
            map[s[right]] = right
            result = max(result, right-left+1)
            right += 1
        return result

import java.util.HashMap;
class Solution {
    public int lengthOfLongestSubstring(String s) {
        HashMap<Character, Integer> map = new HashMap<>();
        int result = 0;
        for (int right=0, left=0; right<s.length(); right++){
            if (map.containsKey(s.charAt(right))){
                left = Math.max(left, map.get(s.charAt(right)) + 1);
            }
            result = Math.max(result, right-left+1);
            map.put(s.charAt(right), right);
        }
        return result;
    }
}