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.
Example 4:
Input: s = "" Output: 0
Constraints:
0 <= s.length <= 5 * 104
s
consists of English letters, digits, symbols and spaces.
class Solution:
def lengthOfLongestSubstring(self, s: str) -> int:
res, chars = 0, dict()
i = j = 0
while j < len(s):
if s[j] in chars:
i = max(i, chars[s[j]] + 1)
res = max(res, j - i + 1)
chars[s[j]] = j
j += 1
return res
class Solution {
public int lengthOfLongestSubstring(String s) {
int res = 0;
Map<Character, Integer> chars = new HashMap<>();
for (int i = 0, j = 0; j < s.length(); ++j) {
char c = s.charAt(j);
if (chars.containsKey(c)) {
i = Math.max(i, chars.get(c) + 1);
}
chars.put(c, j);
res = Math.max(res, j - i + 1);
}
return res;
}
}
function lengthOfLongestSubstring(s: string): number {
// 滑动窗口+哈希表
let left = -1;
let maxLen = 0;
let hashTable = new Map();
for (let right = 0; right < s.length; right++) {
let cur = s.charAt(right);
if (hashTable.has(cur)) {
left = Math.max(left, hashTable.get(cur));
}
hashTable.set(cur, right);
maxLen = Math.max(maxLen, right - left);
}
return maxLen;
};
class Solution {
func lengthOfLongestSubstring(_ s: String) -> Int {
var map = [Character: Int]()
var currentStartingIndex = 0
var i = 0
var maxLength = 0
for char in s {
if map[char] != nil {
if map[char]! >= currentStartingIndex {
maxLength = max(maxLength, i - currentStartingIndex)
currentStartingIndex = map[char]! + 1
}
}
map[char] = i
i += 1
}
return max(maxLength, i - currentStartingIndex)
}
}