题目：
    3.无重复字符的最长子串

    思路：
    滑动窗口
    我用windows.pop(0)代替了left索引

    1. class Solution:
    2.     def lengthOfLongestSubstring(self, s: str) -> int:
    3.         # left = 0
    4.         right = 0
    5.         windows = []
    6.         most_length_child = 0
    7.         while right < len(s):
    8.             windows.append(s[right])
    9.             # print(windows)
    11.             while len(windows) != len(set(windows)):
    12.                 windows.pop(0)
    13.                 # left += 1
    14.             if len(windows) >= most_length_child:
    15.                     most_length_child = len(windows)
    16.             right+= 1
    17.         return most_length_child

    照着模板写，关键是第二个while处的条件

    1. int left = 0, right = 0;
    3. while (right < s.size()) {
    4.     window.add(s[right]);
    5.     right++;
    7.     while (valid) {
    8.         window.remove(s[left]);
    9.         left++;
    10.     }
    11. }

    第二次写的版本

    1. class Solution:
    2.     def lengthOfLongestSubstring(self, s: str) -> int:
    3.         right = 0
    4.         windows = []
    5.         max_substr = 0
    7.         while right < len(s):
    8.             windows.append(s[right])
    9.             right += 1
    11.             while len(windows) != len(set(windows)):
    12.                 windows.pop(0)
    14.             max_substr = max(max_substr, len(windows))
    16.         return max_substr