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125 lines
2.3 KiB
Markdown
125 lines
2.3 KiB
Markdown
# [3. Longest Substring Without Repeating Characters](https://leetcode.com/problems/longest-substring-without-repeating-characters/)
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## 题目
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Given a string, find the length of the longest substring without repeating characters.
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**Example 1**:
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```
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Input: "abcabcbb"
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Output: 3
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Explanation: The answer is "abc", with the length of 3.
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```
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**Example 2**:
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```
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Input: "bbbbb"
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Output: 1
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Explanation: The answer is "b", with the length of 1.
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```
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**Example 3**:
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```
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Input: "pwwkew"
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Output: 3
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Explanation: The answer is "wke", with the length of 3.
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Note that the answer must be a substring, "pwke" is a subsequence and not a substring.
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```
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## 题目大意
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在一个字符串重寻找没有重复字母的最长子串。
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## 解题思路
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这一题和第 438 题，第 3 题，第 76 题，第 567 题类似，用的思想都是"滑动窗口"。
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滑动窗口的右边界不断的右移，只要没有重复的字符，就持续向右扩大窗口边界。一旦出现了重复字符，就需要缩小左边界，直到重复的字符移出了左边界，然后继续移动滑动窗口的右边界。以此类推，每次移动需要计算当前长度，并判断是否需要更新最大长度，最终最大的值就是题目中的所求。
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## 代码
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```go
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package leetcode
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// 解法一 位图
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func lengthOfLongestSubstring(s string) int {
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if len(s) == 0 {
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return 0
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}
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var bitSet [256]bool
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result, left, right := 0, 0, 0
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for left < len(s) {
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// 右侧字符对应的 bitSet 被标记 true，说明此字符在 X 位置重复，需要左侧向前移动，直到将X标记为 false
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if bitSet[s[right]] {
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bitSet[s[left]] = false
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left++
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} else {
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bitSet[s[right]] = true
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right++
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}
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if result < right-left {
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result = right - left
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}
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if left+result >= len(s) || right >= len(s) {
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break
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}
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}
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return result
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}
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// 解法二 滑动窗口
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func lengthOfLongestSubstring_(s string) int {
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if len(s) == 0 {
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return 0
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}
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var freq [256]int
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result, left, right := 0, 0, -1
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for left < len(s) {
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if right+1 < len(s) && freq[s[right+1]-'a'] == 0 {
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freq[s[right+1]-'a']++
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right++
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} else {
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freq[s[left]-'a']--
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left++
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}
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result = max(result, right-left+1)
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}
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return result
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}
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func max(a int, b int) int {
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if a > b {
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return a
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}
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return b
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}
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```
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