Problem Statement:
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 = “dvdf” Output: 3 Explanation: The answer is “vdf”, 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 * 10^4
s
consists of English letters, digits, symbols and spaces.
Solution:
1. HashMap Version
class Solution {
public int lengthOfLongestSubstring(String s) {
int start = 0;
int end = 0;
int maxLength = 0;
HashMap<Character, Integer> positionMap = new HashMap<>();
while (end < s.length()) {
Character ch = s.charAt(end);
/*The use of `Math.max(start, positionMap.get(ch) + 1)` is crucial to
ensure the `start` pointer only moves forward and never backwards.*/
if (positionMap.containsKey(ch)) {
start = Math.max(start, positionMap.get(ch) + 1);
}
positionMap.put(ch, end);
maxLength = Math.max(maxLength, end - start + 1);
end++;
}
return maxLength;
}
}
2. Set Version
public class Solution {
public int lengthOfLongestSubstring(String s) {
int maxLen = 0;
Set<Character> window = new HashSet<>();
int left = 0;
int right = 0;
while(right < s.length()) {
char currentChar = s.charAt(right);
// Contract the window until the duplicate is removed
while (window.contains(currentChar)) {
window.remove(s.charAt(left));
left++;
}
// Add the current character to the window
window.add(currentChar);
// Update the maximum length of the substring
maxLen = Math.max(maxLen, right - left + 1);
right++;
}
return maxLen;
}
}