# 48.最长不含重复字符的子字符串

### 一、 题目

请从字符串中找出一个最长的不包含重复字符的子字符串，计算该最长子字符串的长度。

**示例 1:**

输入: "abcabcbb"

输出: 3

解释: 因为无重复字符的最长子串是 "abc"，所以其长度为 3。

**示例 2:**

输入: "bbbbb"

输出: 1

解释: 因为无重复字符的最长子串是 "b"，所以其长度为 1。

**示例 3:**

输入: "pwwkew"

输出: 3

解释: 因为无重复字符的最长子串是 "wke"，所以其长度为 3。   请注意，你的答案必须是 子串 的长度，"pwke" 是一个子序列，不是子串。  

**提示：**

s.length <= 40000

来源：力扣（LeetCode）

链接：<https://leetcode-cn.com/problems/zui-chang-bu-han-zhong-fu-zi-fu-de-zi-zi-fu-chuan-lcof>

著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。

### 二、 解法

```
func lengthOfLongestSubstring(_ s: String) -> Int {
    guard s.isEmpty == false else {
        return 0
    }
    let stringArr: [Character] = Array(s)
    var maxCount: Int = 0
    /// 最后一次元素出现的下标
    var map: [Character : Int] = [ : ]
    /// 一个重复段内的不重复子串长度
    var temp: Int = 0
    
    for idx in 0..<stringArr.count {
        if let last = map[stringArr[idx]] {
            temp = (temp < (idx - last)) ? (temp + 1) : (idx - last)
        }
        else {
            temp += 1
        }
        map[stringArr[idx]] = idx
        maxCount = max(maxCount, temp)
    }
    
    return maxCount
}
```

可以简化为

```
func lengthOfLongestSubstring(_ s: String) -> Int {
    ...

    for idx in 0..<stringArr.count {
        let last = map[stringArr[idx]] ?? -1
        temp = (temp < (idx - last)) ? (temp + 1) : (idx - last)
        map[stringArr[idx]] = idx
        maxCount = max(maxCount, temp)
    }
    
    ...
}
```


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