題目：
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.
sliding window的題，通過一個set來作爲窗口存放不重複的字符。首先begin和end都指向第一個元素，然後保持begin不動，一個個往右移動end，直到發現一個重複字符，就往右移動begin直到沒有重複的字符，如此循環。寫代碼的時候注意求長度的時候因爲放在end++後面所以不需要end - begin + 1，直接end - begin即可，以及在begin++的時候，別忘了把begin對應的元素從set中移除。
Runtime: 24 ms, faster than 49.44% of C++ online submissions for Longest Substring Without Repeating Characters.
Memory Usage: 13.3 MB, less than 33.83% of C++ online submissions for Longest Substring Without Repeating Characters.
class Solution {
public:
int lengthOfLongestSubstring(string s) {
unordered_set<char> set;
int begin = 0;
int end = 0;
int result = 0;
while (begin <= end && end < s.size()) {
if (set.count(s[end])) {
set.erase(s[begin]); // must erase it from the set
begin++;
}
else {
set.insert(s[end]);
end++;
result = max(result, end - begin); // because end++, so dont' need to + 1
}
}
return result;
}
};
這個寫法還可以通過
