LeetCode #3. Longest Substring Without Repeating Characters C#

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.

Solution:

Use two-pointer and HashTable to solve.

Keep p1 at the beginning of the substring and move p2, add characters to hash table and keep the index as value,

if map contains s[p2] then need to  move p1 to the right of the duplicated char index, then update the index of map[s[p2]] to p2;

also p1 and p2 should only move forward;

ex) "abba" when p1=0, p2 = 2, we met the second ‘b‘ then p1 should move to 2, and p2 stay at 2.

then when p1=2, p2=3 we met the second a, p1 should never move back to index 1 to start over;

Before come up with this solution, my solution was to clear the map and move p1 to the right of the first duplicated char, and same to p2 then go through again and add to map,

This method worst case runtime is O(n2), so exceeded the time Limited

 1 public class Solution {
 2     public int LengthOfLongestSubstring(string s) {
 3         if(string.IsNullOrEmpty(s))
 4         {
 5             return 0;
 6         }
 7         int l = s.Length;
 8         int max = 0;
 9         int p1=0;
10         int p2=0;
11         Dictionary<char,int> map = new Dictionary<char,int>();
12         while(p2<l)
13         {
14             if(!map.ContainsKey(s[p2]))
15             {
16                 map.Add(s[p2], p2);
17             }
18             else
19             {
20                 if(p1<=map[s[p2]])
21                 {
22                    p1=map[s[p2]]+1;
23
24                 }
25                  map[s[p2]]= p2;
26             }
27             max= Math.Max(max, p2-p1+1);
28             p2++;
29
30         }
31         return max;
32     }
33 }
时间: 2024-10-12 20:32:40

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