leetcode-【中等题】3. Longest Substring Without Repeating Characters

本文介绍了一种解决LeetCode上“最长不含重复字符的子字符串”问题的方法。通过跟踪字符最后一次出现的位置,并使用滑动窗口技巧来更新最长子串长度。

题目:

Given a string, find the length of the longest substring without repeating characters.

Examples:

Given "abcabcbb", the answer is "abc", which the length is 3.

Given "bbbbb", the answer is "b", with the length of 1.

Given "pwwkew", 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.

链接https://leetcode.com/problems/longest-substring-without-repeating-characters/

答案:

参考了这个链接后才明白解题思路,然后依据自己的理解写了代码。

大致思路是:从当前字符开始,往前看,没有出现过重复字符的字符串

代码:

 1 #define MAX_LETTER_NUM 255
 2 class Solution {
 3 public:
 4     int lengthOfLongestSubstring(string s) {
 5         if(s.empty())
 6         {
 7             return 0;
 8         }
 9         
10         bool exist[MAX_LETTER_NUM];
11         int position[MAX_LETTER_NUM];
12         int index,rangeIndex;
13         int maxLen = 0;
14         int start = 0;
15         
16         for(index = 0; index < MAX_LETTER_NUM; ++ index)
17         {
18             exist[index] = false;
19             position[index] = 0;
20         }
21         
22         for(index = 0; index < s.size(); ++ index)
23         {
24             if(exist[s[index]])
25             {
26                 for(rangeIndex = start; rangeIndex <= position[s[index]]; ++ rangeIndex)
27                 {
28                     exist[s[rangeIndex]] = false;
29                 }
30                 start = position[s[index]] + 1;
31                 exist[s[index]] = true;
32             }else
33             {
34                 exist[s[index]] = true;
35                 maxLen = (maxLen < index - start + 1) ? index - start + 1:maxLen;
36             }
37             
38             position[s[index]] = index;
39         }
40         
41         return maxLen;
42     }
43 };
View Code

 

转载于:https://www.cnblogs.com/Shirlies/p/5736696.html

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