Description：
Given a string, find the length of the longest substring without repeating characters.
Example 1:
Input: "abcabcbb"
Output: 3
Explanation: The answer is "abc", with the length of 3.
Example 2:
Input: "bbbbb"
Output: 1
Explanation: The answer is "b", with the length of 1.
Example 3:
Input: "pwwkew"
Output: 3
Explanation: 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.
翻譯：
給定一個字符串，找出其中不含有重複字符的 最長子串 的長度。
Kotlin代碼實現①
public int lengthOfLongestSubstring(String s) {
Map<Character, Integer> map = new HashMap<Character, Integer>();
char[] charArr = s.toCharArray();
int max = 0;
int count = 0;
for (int i = 0; i < charArr.length; i++) {
Integer last = map.get(charArr[i]);
if (last == null) {
count++;
if (count > max) {
max = count;
}
} else {
int start=i-count;
for(int j=start;j<last;j++){
map.remove(charArr[j]);
count--;
}
}
map.put(charArr[i], i);
}
return max;
}
Kotlin代碼實現②
fun lengthOfLongestSubstring(s: String): Int {
val n = s.length
val set = HashSet<Char>()
var max = 0
var i = 0
var j = 0
while (i < n && j < n) {
if (!set.contains(s[j])) {
set.add(s[j++])
max = Math.max(max, j - i)
} else {
set.remove(s[i++])
}
}
return max
}
參考：https://leetcode.com/problems/longest-substring-without-repeating-characters