Longest Substring Without Repeating Characters

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.

记一个满有趣的解法

首先是我的暴力解法= =

 1 class Solution {
 2 public:
 3     int lengthOfLongestSubstring(string s) {
 4         if(s.empty())
 5             return 0;
 6         int length = 1;   
 7         int len = 1;
 8         for(int i = 1; i < s.size(); i++)
 9         {
10             if(s[i] != s[i-1])
11             {
12                 int j = i - 2;
13                 int n = len - 1;
14                 while(n)
15                 {
16                     if(s[j] == s[i])
17                     {
18                         len = i - j;
19                         break;
20                     } else
21                     {
22                         n--;
23                         j--;
24                     }
25                 }
26                 if(j == i - len - 1)
27                 {
28                     len++;
29                     length = max(length, len);
30                 }
31             } else
32                 len = 1;
33         }
34         return length;
35     }
36 };

 

别人的时间复杂度为O(n)的解法...

思想是利用dict[]记录出现过的字符在s中的下标。如果 if(dict[s[i]] > start) 为真，则说明字符再次出现，此时用start记录前一次出现时的下标，然后更新dict[]中该字符的值。可以看到越先出现字符lict[]的值越小，因为i是递增的，这保证了当有重复的字符出现时，

dict[s[i]] 必然大于 start，所以start记录的肯定是最后出现的重复字符的前一次出现时的下标。还是看代码吧，说着太绕了。

 1 int lengthOfLongestSubstring(string s) {
 2         vector<int> dict(256, -1);
 3         int maxLen = 0, start = -1;
 4         for (int i = 0; i != s.length(); i++) {
 5             if (dict[s[i]] > start)
 6                 start = dict[s[i]];
 7             dict[s[i]] = i;
 8             maxLen = max(maxLen, i - start);
 9         }
10         return maxLen;
11 }

 

最近在leetcode刷题，遇到有意思的才会更新到博客，所以最近都没怎么更新= =

posted @ 2018-07-05 03:47  小小Cv  阅读(114)  评论(0)  编辑  收藏  举报