add Longest Substring Without Repeating Characters solution

Author: HoonDongKang

longest-substring-without-repeating-characters/HoonDongKang.ts

@@ -0,0 +1,80 @@
+/**
+ * [Problem]: [3] Longest Substring Without Repeating Characters
+ * (https://leetcode.com/problems/longest-substring-without-repeating-characters/description/)
+ */
+
+/*
+ * 1. 중복없이 이어는 문자열 중 가장 긴 문자열
+ */
+function lengthOfLongestSubstring(s: string): number {
+    // 시간 복잡도: O(n^2)
+    // 공간 복잡도: O(n)
+    function bruteForceFunc(s: string): number {
+        let max = 0;
+        for (let i = 0; i < s.length; i++) {
+            const duplicate = new Set<string>();
+            let j = i;
+            let count = 0;
+
+            while (j < s.length && !duplicate.has(s[j])) {
+                duplicate.add(s[j]);
+                count++;
+                j++;
+            }
+
+            max = Math.max(max, count);
+        }
+
+        return max;
+    }
+    //시간복잡도 O(n)
+    //공간복잡도 O(n)
+    function slidingWindowFunc(s: string): number {
+        let left = 0;
+        let right = 0;
+        let window = new Set<string>();
+        let max = 0;
+        while (right < s.length) {
+            if (window.has(s[right])) {
+                window.delete(s[left]);
+                left++;
+            } else {
+                window.add(s[right]);
+                max = Math.max(max, right - left + 1);
+                right++;
+            }
+            // while (window.has(s[right])) {
+            //     window.delete(s[left]);
+            //     left++;
+            // }
+
+            // window.add(s[right]);
+            // max = Math.max(max, right - left + 1);
+            // right++;
+        }
+
+        return max;
+    }
+
+    //시간복잡도 O(n)
+    //공간복잡도 O(n)
+    //set에서 left 인덱스를 조금 더 빠르게 가져올 수 있음
+    function mapFunc(s: string): number {
+        let left = 0;
+        let max = 0;
+        let map = new Map<string, number>();
+
+        for (let right = 0; right < s.length; right++) {
+            if (map.has(s[right])) {
+                left = Math.max(left, map.get(s[right])! + 1);
+            }
+
+            map.set(s[right], right);
+            max = Math.max(max, right - left + 1);
+        }
+
+        return max;
+    }
+
+    return mapFunc(s);
+}