feat: add solutions to lcof problem: No.48

lcof/面试题48. 最长不含重复字符的子字符串/README.md

@@ -40,13 +40,15 @@
 
 ## 解法
 
-“滑动窗口 + 哈希表”。
+**方法一：双指针 + 哈希表**
 
-定义一个哈希表记录当前窗口内出现的字符，i、j 分别表示不重复子串的结束位置和开始位置，res 表示无重复字符子串的最大长度。
+我们用双指针 $j$ 和 $i$ 分别表示子串的左右边界，其中 $j$ 是滑动窗口的左边界，$i$ 是滑动窗口的右边界，用哈希表 $vis$ 记录每个字符是否出现过。
 
-遍历 i，若 `[j, i - 1]` 窗口内存在 `s[i]`，则 j 循环向右移动，更新哈希表，直至 `[j, i - 1]` 窗口不存在 `s[i]`，循环结束。将 `s[i]` 加入哈希表中，此时 `[j, i]` 窗口内不含重复元素，更新 res 的最大值：`res = max(res, i - j + 1)`。
+遍历字符串 $s$，如果此时 $s[i]$ 在哈希表 $vis$ 中存在，说明 $s[i]$ 重复了，我们需要将左边界 $j$ 右移，直到 $s[i]$ 不在哈希表 $vis$ 中为止，然后将 $s[i]$ 加入哈希表 $vis$ 中。此时，我们更新无重复字符子串的最大长度，即 $ans = max(ans, i - j + 1)$。
 
-最后返回 res 即可。
+遍历结束后，我们返回 $ans$ 即可。
+
+时间复杂度 $O(n)$，空间复杂度 $O(C)$。其中 $n$ 是字符串 $s$ 的长度；而 $C$ 是字符集的大小。
 
 <!-- tabs:start -->
 
@@ -55,35 +57,46 @@
 ```python
 class Solution:
     def lengthOfLongestSubstring(self, s: str) -> int:
-        i = j = res = 0
-        chars = set()
-        while i < len(s):
-            while s[i] in chars:
-                if s[j] in chars:
-                    chars.remove(s[j])
+        cnt = Counter()
+        ans = j = 0
+        for i, c in enumerate(s):
+            cnt[c] += 1
+            while cnt[c] > 1:
+                cnt[s[j]] -= 1
                 j += 1
-            chars.add(s[i])
-            res = max(res, i - j + 1)
-            i += 1
-        return res
+            ans = max(ans, i - j + 1)
+        return ans
+```
+
+```python
+class Solution:
+    def lengthOfLongestSubstring(self, s: str) -> int:
+        vis = set()
+        ans = j = 0
+        for i, c in enumerate(s):
+            while c in vis:
+                vis.remove(s[j])
+                j += 1
+            vis.add(c)
+            ans = max(ans, i - j + 1)
+        return ans
 ```
 
 ### **Java**
 
 ```java
 class Solution {
     public int lengthOfLongestSubstring(String s) {
-        int res = 0;
-        Set<Character> set = new HashSet<>();
-        for (int i = 0, j = 0; i < s.length(); ++i) {
-            char c = s.charAt(i);
-            while (set.contains(c)) {
-                set.remove(s.charAt(j++));
+        int ans = 0, j = 0;
+        Set<Character> vis = new HashSet<>();
+        for (int i = 0; i < s.length(); ++i) {
+            while (vis.contains(s.charAt(i))) {
+                vis.remove(s.charAt(j++));
             }
-            set.add(c);
-            res = Math.max(res, i - j + 1);
+            vis.add(s.charAt(i));
+            ans = Math.max(ans, i - j + 1);
         }
-        return res;
+        return ans;
     }
 }
 ```
@@ -94,35 +107,35 @@ class Solution {
 class Solution {
 public:
     int lengthOfLongestSubstring(string s) {
-        int res = 0;
-        unordered_set<char> chars;
+        int ans = 0;
+        unordered_set<char> vis;
         for (int i = 0, j = 0; i < s.size(); ++i) {
-            while (chars.count(s[i])) {
-                chars.erase(s[j++]);
+            while (vis.count(s[i])) {
+                vis.erase(s[j++]);
             }
-            chars.insert(s[i]);
-            res = max(res, i - j + 1);
+            vis.insert(s[i]);
+            ans = max(ans, i - j + 1);
         }
-        return res;
+        return ans;
     }
 };
 ```
 
 ### **Go**
 
 ```go
-func lengthOfLongestSubstring(s string) int {
-	chars := make(map[byte]bool)
-	res := 0
-	for i, j := 0, 0; i < len(s); i++ {
-		for chars[s[i]] {
-			chars[s[j]] = false
+func lengthOfLongestSubstring(s string) (ans int) {
+	vis := map[byte]bool{}
+	j := 0
+	for i := range s {
+		for vis[s[i]] {
+			vis[s[j]] = false
 			j++
 		}
-		chars[s[i]] = true
-		res = max(res, i-j+1)
+		vis[s[i]] = true
+		ans = max(ans, i-j+1)
 	}
-	return res
+	return
 }
 
 func max(a, b int) int {
@@ -141,53 +154,33 @@ func max(a, b int) int {
  * @return {number}
  */
 var lengthOfLongestSubstring = function (s) {
-    let res = 0;
-    let chars = new Set();
+    let ans = 0;
+    let vis = new Set();
     for (let i = 0, j = 0; i < s.length; ++i) {
-        while (chars.has(s[i])) {
-            chars.delete(s[j++]);
+        while (vis.has(s[i])) {
+            vis.delete(s[j++]);
         }
-        chars.add(s[i]);
-        res = Math.max(res, i - j + 1);
+        vis.add(s[i]);
+        ans = Math.max(ans, i - j + 1);
     }
-    return res;
+    return ans;
 };
 ```
 
 ### **TypeScript**
 
 ```ts
 function lengthOfLongestSubstring(s: string): number {
-    const n = s.length;
-    const set = new Set<string>();
-    let res = 0;
-    let i = 0;
-    for (let j = 0; j < n; j++) {
-        const c = s[j];
-        while (set.has(c)) {
-            set.delete(s[i++]);
-        }
-        set.add(c);
-        res = Math.max(res, set.size);
-    }
-    return res;
-}
-```
-
-```ts
-function lengthOfLongestSubstring(s: string): number {
-    const map = new Map<string, number>();
-    const n = s.length;
-    let res = 0;
-    let i = -1;
-    for (let j = 0; j < n; j++) {
-        if (map.has(s[j])) {
-            i = Math.max(i, map.get(s[j]));
+    let ans = 0;
+    let vis = new Set<string>();
+    for (let i = 0, j = 0; i < s.length; ++i) {
+        while (vis.has(s[i])) {
+            vis.delete(s[j++]);
         }
-        map.set(s[j], j);
-        res = Math.max(res, j - i);
+        vis.add(s[i]);
+        ans = Math.max(ans, i - j + 1);
     }
-    return res;
+    return ans;
 }
 ```
 
@@ -243,15 +236,14 @@ impl Solution {
 ```cs
 public class Solution {
     public int LengthOfLongestSubstring(string s) {
-        var set = new HashSet<char>();
+        var vis = new HashSet<char>();
         int ans = 0;
-        for (int l=0, r=0; r < s.Length; r++) {
-            while (set.Contains(s[r]))
-            {
-                set.Remove(s[l++]);
+        for (int i = 0, j = 0; i < s.Length; ++i) {
+            while (vis.Contains(s[i])) {
+                vis.Remove(s[j++]);
             }
-            ans = Math.Max(r - l + 1, ans);
-            set.Add(s[r]);
+            vis.Add(s[i]);
+            ans = Math.Max(ans, i - j + 1);
         }
         return ans;
     }

lcof/面试题48. 最长不含重复字符的子字符串/Solution.cpp

@@ -1,15 +1,15 @@
 class Solution {
 public:
     int lengthOfLongestSubstring(string s) {
-        int res = 0;
-        unordered_set<char> chars;
+        int ans = 0;
+        unordered_set<char> vis;
         for (int i = 0, j = 0; i < s.size(); ++i) {
-            while (chars.count(s[i])) {
-                chars.erase(s[j++]);
+            while (vis.count(s[i])) {
+                vis.erase(s[j++]);
             }
-            chars.insert(s[i]);
-            res = max(res, i - j + 1);
+            vis.insert(s[i]);
+            ans = max(ans, i - j + 1);
         }
-        return res;
+        return ans;
     }
 };

lcof/面试题48. 最长不含重复字符的子字符串/Solution.cs

@@ -1,14 +1,13 @@
 public class Solution {
     public int LengthOfLongestSubstring(string s) {
-        var set = new HashSet<char>();
+        var vis = new HashSet<char>();
         int ans = 0;
-        for (int l=0, r=0; r < s.Length; r++) {
-            while (set.Contains(s[r]))
-            {
-                set.Remove(s[l++]);
+        for (int i = 0, j = 0; i < s.Length; ++i) {
+            while (vis.Contains(s[i])) {
+                vis.Remove(s[j++]);
             }
-            ans = Math.Max(r - l + 1, ans);
-            set.Add(s[r]);
+            vis.Add(s[i]);
+            ans = Math.Max(ans, i - j + 1);
         }
         return ans;
     }

lcof/面试题48. 最长不含重复字符的子字符串/Solution.go

@@ -1,15 +1,15 @@
-func lengthOfLongestSubstring(s string) int {
-	chars := make(map[byte]bool)
-	res := 0
-	for i, j := 0, 0; i < len(s); i++ {
-		for chars[s[i]] {
-			chars[s[j]] = false
+func lengthOfLongestSubstring(s string) (ans int) {
+	vis := map[byte]bool{}
+	j := 0
+	for i := range s {
+		for vis[s[i]] {
+			vis[s[j]] = false
 			j++
 		}
-		chars[s[i]] = true
-		res = max(res, i-j+1)
+		vis[s[i]] = true
+		ans = max(ans, i-j+1)
 	}
-	return res
+	return
 }
 
 func max(a, b int) int {

lcof/面试题48. 最长不含重复字符的子字符串/Solution.java

@@ -1,15 +1,14 @@
 class Solution {
     public int lengthOfLongestSubstring(String s) {
-        int res = 0;
-        Set<Character> set = new HashSet<>();
-        for (int i = 0, j = 0; i < s.length(); ++i) {
-            char c = s.charAt(i);
-            while (set.contains(c)) {
-                set.remove(s.charAt(j++));
+        int ans = 0, j = 0;
+        Set<Character> vis = new HashSet<>();
+        for (int i = 0; i < s.length(); ++i) {
+            while (vis.contains(s.charAt(i))) {
+                vis.remove(s.charAt(j++));
             }
-            set.add(c);
-            res = Math.max(res, i - j + 1);
+            vis.add(s.charAt(i));
+            ans = Math.max(ans, i - j + 1);
         }
-        return res;
+        return ans;
     }
 }

lcof/面试题48. 最长不含重复字符的子字符串/Solution.js

@@ -3,14 +3,14 @@
  * @return {number}
  */
 var lengthOfLongestSubstring = function (s) {
-    let res = 0;
-    let chars = new Set();
+    let ans = 0;
+    let vis = new Set();
     for (let i = 0, j = 0; i < s.length; ++i) {
-        while (chars.has(s[i])) {
-            chars.delete(s[j++]);
+        while (vis.has(s[i])) {
+            vis.delete(s[j++]);
         }
-        chars.add(s[i]);
-        res = Math.max(res, i - j + 1);
+        vis.add(s[i]);
+        ans = Math.max(ans, i - j + 1);
     }
-    return res;
+    return ans;
 };

lcof/面试题48. 最长不含重复字符的子字符串/Solution.py

@@ -1,13 +1,11 @@
 class Solution:
     def lengthOfLongestSubstring(self, s: str) -> int:
-        i = j = res = 0
-        chars = set()
-        while i < len(s):
-            while s[i] in chars:
-                if s[j] in chars:
-                    chars.remove(s[j])
+        vis = set()
+        ans = j = 0
+        for i, c in enumerate(s):
+            while c in vis:
+                vis.remove(s[j])
                 j += 1
-            chars.add(s[i])
-            res = max(res, i - j + 1)
-            i += 1
-        return res
+            vis.add(c)
+            ans = max(ans, i - j + 1)
+        return ans

lcof/面试题48. 最长不含重复字符的子字符串/Solution.ts

@@ -1,15 +1,12 @@
 function lengthOfLongestSubstring(s: string): number {
-    const n = s.length;
-    const set = new Set<string>();
-    let res = 0;
-    let i = 0;
-    for (let j = 0; j < n; j++) {
-        const c = s[j];
-        while (set.has(c)) {
-            set.delete(s[i++]);
+    let ans = 0;
+    let vis = new Set<string>();
+    for (let i = 0, j = 0; i < s.length; ++i) {
+        while (vis.has(s[i])) {
+            vis.delete(s[j++]);
         }
-        set.add(c);
-        res = Math.max(res, set.size);
+        vis.add(s[i]);
+        ans = Math.max(ans, i - j + 1);
     }
-    return res;
+    return ans;
 }