[add] LeetCode 3. Longest Substring Without Repeating Characters

Author: brianway

algorithms-leetcode/README.md

@@ -29,6 +29,7 @@ Leetcode problems classified by company:
 |:----|:----|:----|:----|:----|
 |1|Two Sum|Easy|Hash Table||
 |2|Add Two Numbers|Medium|Linked List||
+|3|Longest Substring Without Repeating Characters|Medium|Sliding Window,Hash Table||
 |5|Longest Palindromic Substring|Medium|Dynamic Programming/Two Pointers|一题多解|
 |6|Zigzag Conversion|Medium||一题多解TODO|
 |10|Regular Expression Matching|Hard|||

algorithms-leetcode/src/main/java/com/brianway/learning/algorithms/leetcode/medium/LongestSubstringWithoutRepeatingCharacters.java

@@ -0,0 +1,48 @@
+package com.brianway.learning.algorithms.leetcode.medium;
+
+import java.util.HashMap;
+import java.util.Map;
+
+/**
+ * LeetCode 3. Longest Substring Without Repeating Characters
+ * Question: https://leetcode.com/problems/longest-substring-without-repeating-characters/
+ * 关键题设:  无
+ * * @auther brian
+ *
+ * @since 2022/11/30 23:37
+ */
+public class LongestSubstringWithoutRepeatingCharacters {
+
+    public int lengthOfLongestSubstring(String s) {
+        return 0;
+    }
+
+    /**
+     * 滑动窗口+hash表
+     *
+     * 测试用例："tmmzuxt"
+     */
+    public class LongestSubstringWithoutRepeatingCharacters0 extends LongestSubstringWithoutRepeatingCharacters {
+        @Override
+        public int lengthOfLongestSubstring(String s) {
+            char[] chars = s.toCharArray();
+            Map<Character, Integer> exist = new HashMap<>();
+            int maxLen = 0;
+            int start = 0;
+
+            for (int i = 0; i < chars.length; i++) {
+                Integer lastIndex = exist.get(chars[i]);
+                if (lastIndex == null || lastIndex < start) {
+                    // 未出现过，或者出现在左指针之前，直接计算length
+                    maxLen = Math.max(i - start + 1, maxLen);
+                } else {
+                    // 当前chars[i]出现在start后，则需要右移start
+                    start = lastIndex + 1;
+                }
+                exist.put(chars[i], i);
+            }
+            return maxLen;
+        }
+    }
+
+}