LeetCode刷题系列 - 003题Longest Substring Without Repeating Characters

Posted 2021-10-25 xuelisheng

题目：

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.

代码(C++实现)：

 1 class Solution {
 2 public:
 3     int lengthOfLongestSubstring(string s) 
 4     {
 5         // 定义一个map用来存放整个字符串s
 6         unordered_map<char, int> unmap;
 7         
 8         // tempLength记录每次扫描位置开始的一次统计的最长字符串的长度
 9         int tempLength = 0;
10         // 众多tempLength中的最大值
11         int maxLength = 0;
12         
13         // 第一层循环：分别以s中的每个字符为基准，进行遍历
14         for (int j = 0; j < s.size(); j++)
15         {
16             // 第二层循环：以当前第一层循环中当前的字符为基准，进行遍历，统计以此字符为基准的tempLength
17             for (int i = j; i < s.size(); i++)
18             {
19                 // 是否tempLength继续增加的条件是，map中没有出现过当前指向的字符
20                 if (unmap.count(s[i]) == 0)                
21                 {
22                     pair<char, int> myshopping(s[i], i);
23                     // 如果当前的map中无此字符，将当前字符插入到map中
24                     unmap.insert(myshopping);
25                     tempLength++;
26                     maxLength = maxLength > tempLength ? maxLength : tempLength;
27                 }
28                 // 当前字符已经在map中了，直接break，并将本次使用的map进行清除操作
29                 else
30                 {
31                     
32                     tempLength = 0;
33                     unmap.clear();
34                     break;
35                 }
36 
37             }
38         }
39     
40         return maxLength;
41         }
42 };