feat: 3. 无重复字符的最长子串

Author: vesper

README.md

@@ -21,6 +21,7 @@
     - [731. 我的日程安排表 II - Medium](./src/leetcode/misc/leet_zh_731/MyCalendarTwo.java)
     - [732. 我的日程安排表 III - Hard](./src/leetcode/misc/leet_zh_732/MyCalendarThree.java)
 - 【字符串】
+    - [3. 无重复字符的最长子串 - Medium](src/leetcode/string/leet_zh_3/Solution.java)
     - [76. 最小覆盖子串 - Hard](src/leetcode/string/leet_zh_76/Solution.java)
     - [438. 找到字符串中所有字母异位词 - Medium](src/leetcode/string/leet_zh_438/Solution.java)
     - [无重复字符的最长子串 - Medium](src/leetcode_specialized/bytedance/string/longest_substring_without_repeating_characters/Solution.java)

src/leetcode/string/leet_zh_3/Solution.java

@@ -0,0 +1,72 @@
+package leetcode.string.leet_zh_3;
+
+import java.util.HashMap;
+import java.util.Map;
+
+public class Solution {
+    /*
+     * Title: 3. 无重复字符的最长子串
+     * Link : https://leetcode-cn.com/problems/longest-substring-without-repeating-characters
+     * Label: ["哈希表", "双指针", "字符串", "Sliding Window"]
+     * Diff : Medium
+     * Desc :
+     * 给定一个字符串，请你找出其中不含有重复字符的 最长子串 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。
+
+     * 执行用时 : 9 ms
+     * 内存消耗 : 39.4 MB
+     * */
+
+    public int lengthOfLongestSubstring(String s) {
+        int maxLength = 0;
+        if (s.length() == 0) {
+            return maxLength;
+        }
+        char c;
+        int left = 0, right = 0;
+        Map<Character, Integer> windows = new HashMap<>();
+        while (right < s.length()) {
+            c = s.charAt(right);
+            if (!windows.containsKey(c)) {
+                windows.put(c, right);
+                if (maxLength < right - left + 1) {
+                    maxLength = right - left + 1;
+                }
+            } else {
+                int index = windows.get(c) + 1;
+                for (int i = left; i < index; i++) {
+                    char c2 = s.charAt(i);
+                    windows.remove(c2);
+                }
+                left = index;
+                windows.put(c, right);
+            }
+            right++;
+        }
+        return maxLength;
+    }
+
+    public static void main(String[] args) {
+        Solution ss = new Solution();
+//        System.out.println(ss.lengthOfLongestSubstring("abcabcbb"));
+//        System.out.println(ss.lengthOfLongestSubstring("bbbbb"));
+//        System.out.println(ss.lengthOfLongestSubstring("pwwkew"));
+        System.out.println(ss.lengthOfLongestSubstring("tmmzuxt1z"));
+    }
+}