Longest Substring Without Repeating Characters - LeetCode
Given a string s, find the length of the longest substring without repeating characters.
Example 1: Input: s = “abcabcbb” Output: 3 Explanation: The answer is “abc”, with the length of 3.
Example 2: Input: s = “bbbbb” Output: 1 Explanation: The answer is “b”, with the length of 1.
Example 3: Input: s = “pwwkew” Output: 3 Explanation: The answer is “wke”, with the length of 3. Notice that the answer must be a substring, “pwke” is a subsequence and not a substring.
Constraints:
0 <= s.length <= 5 * 104
s consists of English letters, digits, symbols and spaces.
- code
class Solution:
def lengthOfLongestSubstring(self, s: str) -> int:
res = 0
sub = []
for i,v in enumerate(s):
if v not in sub:
sub.append(v)
res = max(res, len(sub))
else:
cur_v_i = sub.index(v)
sub = sub[cur_v_i+1:]
sub.append(v)
return res
- code
class Solution {
public int lengthOfLongestSubstring(String s) {
int l = 0;
int res = 0;
Set<Character> set = new HashSet<>();
for (int r = 0; r < s.length(); r++){
if (!set.contains(s.charAt(r))){
set.add(s.charAt(r));
res = Math.max(res, r - l + 1);
continue;
}
while (set.contains(s.charAt(r))){
set.remove(s.charAt(l));
l++;
}
set.add(s.charAt(r));oo
}
return res;
}
}
- code
class Solution {
public int lengthOfLongestSubstring(String s) {
int l = 0;
int res = 0;
Map<Character,Integer> map = new HashMap<>();
for (int r = 0; r < s.length(); r++){
if (map.containsKey(s.charAt(r)) && l <= map.get(s.charAt(r))){
l = map.get(s.charAt(r)) + 1;
}else{
res = Math.max(res, r - l + 1);
}
map.put(s.charAt(r), r);
}
return res;
}
}
- code #slidingwindow
class Solution:
# @return an integer
def lengthOfLongestSubstring(self, s):
start = maxLength = 0
usedChar = {}
for i in range(len(s)):
if s[i] in usedChar and start <= usedChar[s[i]]:
start = usedChar[s[i]] + 1
else:
maxLength = max(maxLength, i - start + 1)
usedChar[s[i]] = i
return maxLength