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# 3. Longest Substring Without Repeating Characters

Hash table

## 1.問題

* 給予一個string, 找出最長的沒有重複字元的substring

![](https://901207480-files.gitbook.io/~/files/v0/b/gitbook-legacy-files/o/assets%2F-LGKoChvN9am4__HCIRK%2F-LJDEYcFwJuU7Pa6wKko%2F-LJDIcBYkbN81ovtqFDe%2F2018080601.jpg?alt=media\&token=1081f518-9e3f-4d65-9383-7e5e82f2bb9a)

## 2.想法

* 用vector來記錄曾經碰到過的字元, 一旦碰到第二次則將從start開始的紀錄取消並且寫入新的紀錄, 移動start, 紀錄長度為start到當前位置的長度
* 用vector代替map存放字母出現次數的好處是: 將會依照字母順序
* 對字串中每個子母進行標記: 出現兩次則清空, 並將first移到第一個重複字的下一個字母

## 3.程式碼

```
class Solution {
public:
    int lengthOfLongestSubstring(string s) {
        vector<int> cnt(128, 0);
        int first = 0, size = s.size(), maxSize = 0;
        for (int i = 0; i < size; i++) {
            cnt[s[i]]++;
            if (cnt[s[i]] > 1) {
                //重要: 將first移動到第一個重複字出現的下一個字母
                for (; s[first] != s[i]; first++) {
                    cnt[s[first]] = 0;
                }
                first++;
            }
            maxSize = max(i - first + 1, maxSize); 
        } 
        return maxSize;
    }
};
```

## 4.Performance

![](https://901207480-files.gitbook.io/~/files/v0/b/gitbook-legacy-files/o/assets%2F-LGKoChvN9am4__HCIRK%2F-LJDEYcFwJuU7Pa6wKko%2F-LJDItV1yWwdnNcMdC6J%2F2018080602.jpg?alt=media\&token=74cc7031-e474-4cd8-a64f-d99ad40563a9)