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.
题意简单明了大致就是计算一个字符串的无重复字符 的最长子字符串长度（子字符串即为连续的子字符串）
最容易想到的实现就是两层for循环遍历，找到最大子字符串，下面是笔者第一版的实现
public int lengthOfLongestSubstring(String s) {
int maxlength = 0;
for (int i = 0; i < s.length(); i++) {
Set<Character> numbers = new HashSet<>();
int times = 0;
for (int j = i; j < s.length(); j++) {
numbers.add(s.charAt(j));
times++;
int size = numbers.size();
if (size < times || size >= s.length() - i) {
if (maxlength < size) {
maxlength = size;
}
break;
}
}
}
return maxlength;
}
提交后发现：
Runtime: 54 ms, faster than 22.41% of Java online submissions for Longest Substring Without Repeating Characters.
Memory Usage: 38.5 MB, less than 31.95% of Java online submissions for Longest Substring Without Repeating Characters.
这一版的算法复杂度O（n^2），但是结果被leetcode大神完虐。这让笔者意识到这道题应该有线性解或者对数解。受限于目前算法的掌握程度还没有学习到字符串算法哪里，但是又不想去参考答案，所以这一版先到此告一段落。等笔者学习完字符串的各种 算法后再回过头来重新实现，所以标题为(defect) 。
