feat: add solutions to lc problems: No.0369,0387,0434

solution/0300-0399/0369.Plus One Linked List/README.md

@@ -57,13 +57,13 @@ tags:
 
 ### 方法一：链表遍历
 
-我们先设置一个虚拟头节点 `dummy`，初始值为 $0$，指向链表头节点 `head`。
+我们先设置一个虚拟头节点 $\textit{dummy}$，初始时 $\textit{dummy}$ 的值为 $0$，并且 $\textit{dummy}$ 的后继节点为链表 $\textit{head}$。
 
-然后从链表头节点开始遍历，找出链表最后一个值不等于 $9$ 的节点 `target`，将 `target` 的值加 $1$。接着将 `target` 之后的所有节点值置为 $0$。
+接下来，我们从虚拟头节点开始遍历链表，找到最后一个不为 $9$ 的节点，将其值加 $1$，并将该节点之后的所有节点的值置为 $0$。
 
-需要注意的是，如果链表中所有节点值都为 $9$，那么遍历结束后，`target` 会指向空节点，这时我们需要将 `dummy` 的值加 $1$，然后返回 `dummy`，否则返回 `dummy` 的下一个节点。
+最后，我们判断虚拟头节点的值是否为 $1$，如果为 $1$，则返回 $\textit{dummy}$，否则返回 $\textit{dummy}$ 的后继节点。
 
-时间复杂度 $O(n)$，空间复杂度 $O(1)$。其中 $n$ 为链表的长度。
+时间复杂度 $O(n)$，其中 $n$ 是链表的长度。空间复杂度 $O(1)$。
 
 <!-- tabs:start -->
 
@@ -76,7 +76,7 @@ tags:
 #         self.val = val
 #         self.next = next
 class Solution:
-    def plusOne(self, head: ListNode) -> ListNode:
+    def plusOne(self, head: Optional[ListNode]) -> Optional[ListNode]:
         dummy = ListNode(0, head)
         target = dummy
         while head:
@@ -143,17 +143,16 @@ public:
     ListNode* plusOne(ListNode* head) {
         ListNode* dummy = new ListNode(0, head);
         ListNode* target = dummy;
-        while (head) {
-            if (head->val != 9) target = head;
-            head = head->next;
+        for (; head; head = head->next) {
+            if (head->val != 9) {
+                target = head;
+            }
         }
-        ++target->val;
-        target = target->next;
-        while (target) {
+        target->val++;
+        for (target = target->next; target; target = target->next) {
             target->val = 0;
-            target = target->next;
         }
-        return dummy->val == 1 ? dummy : dummy->next;
+        return dummy->val ? dummy : dummy->next;
     }
 };
 ```
@@ -178,10 +177,8 @@ func plusOne(head *ListNode) *ListNode {
 		head = head.Next
 	}
 	target.Val++
-	target = target.Next
-	for target != nil {
+	for target = target.Next; target != nil; target = target.Next {
 		target.Val = 0
-		target = target.Next
 	}
 	if dummy.Val == 1 {
 		return dummy
@@ -190,6 +187,38 @@ func plusOne(head *ListNode) *ListNode {
 }
 ```
 
+#### TypeScript
+
+```ts
+/**
+ * Definition for singly-linked list.
+ * class ListNode {
+ *     val: number
+ *     next: ListNode | null
+ *     constructor(val?: number, next?: ListNode | null) {
+ *         this.val = (val===undefined ? 0 : val)
+ *         this.next = (next===undefined ? null : next)
+ *     }
+ * }
+ */
+
+function plusOne(head: ListNode | null): ListNode | null {
+    const dummy = new ListNode(0, head);
+    let target = dummy;
+    while (head) {
+        if (head.val !== 9) {
+            target = head;
+        }
+        head = head.next;
+    }
+    target.val++;
+    for (target = target.next; target; target = target.next) {
+        target.val = 0;
+    }
+    return dummy.val ? dummy : dummy.next;
+}
+```
+
 <!-- tabs:end -->
 
 <!-- solution:end -->

solution/0300-0399/0369.Plus One Linked List/README_EN.md

@@ -44,7 +44,15 @@ tags:
 
 <!-- solution:start -->
 
-### Solution 1
+### Solution 1: Linked List Traversal
+
+We first set a dummy head node $\textit{dummy}$, initially with a value of $0$, and the successor node of $\textit{dummy}$ is the linked list $\textit{head}$.
+
+Next, we traverse the linked list starting from the dummy head node, find the last node that is not $9$, increment its value by $1$, and set the values of all nodes after this node to $0$.
+
+Finally, we check if the value of the dummy head node is $1$. If it is $1$, we return $\textit{dummy}$; otherwise, we return the successor node of $\textit{dummy}$.
+
+The time complexity is $O(n)$, where $n$ is the length of the linked list. The space complexity is $O(1)$.
 
 <!-- tabs:start -->
 
@@ -57,7 +65,7 @@ tags:
 #         self.val = val
 #         self.next = next
 class Solution:
-    def plusOne(self, head: ListNode) -> ListNode:
+    def plusOne(self, head: Optional[ListNode]) -> Optional[ListNode]:
         dummy = ListNode(0, head)
         target = dummy
         while head:
@@ -124,17 +132,16 @@ public:
     ListNode* plusOne(ListNode* head) {
         ListNode* dummy = new ListNode(0, head);
         ListNode* target = dummy;
-        while (head) {
-            if (head->val != 9) target = head;
-            head = head->next;
+        for (; head; head = head->next) {
+            if (head->val != 9) {
+                target = head;
+            }
         }
-        ++target->val;
-        target = target->next;
-        while (target) {
+        target->val++;
+        for (target = target->next; target; target = target->next) {
             target->val = 0;
-            target = target->next;
         }
-        return dummy->val == 1 ? dummy : dummy->next;
+        return dummy->val ? dummy : dummy->next;
     }
 };
 ```
@@ -159,10 +166,8 @@ func plusOne(head *ListNode) *ListNode {
 		head = head.Next
 	}
 	target.Val++
-	target = target.Next
-	for target != nil {
+	for target = target.Next; target != nil; target = target.Next {
 		target.Val = 0
-		target = target.Next
 	}
 	if dummy.Val == 1 {
 		return dummy
@@ -171,6 +176,38 @@ func plusOne(head *ListNode) *ListNode {
 }
 ```
 
+#### TypeScript
+
+```ts
+/**
+ * Definition for singly-linked list.
+ * class ListNode {
+ *     val: number
+ *     next: ListNode | null
+ *     constructor(val?: number, next?: ListNode | null) {
+ *         this.val = (val===undefined ? 0 : val)
+ *         this.next = (next===undefined ? null : next)
+ *     }
+ * }
+ */
+
+function plusOne(head: ListNode | null): ListNode | null {
+    const dummy = new ListNode(0, head);
+    let target = dummy;
+    while (head) {
+        if (head.val !== 9) {
+            target = head;
+        }
+        head = head.next;
+    }
+    target.val++;
+    for (target = target.next; target; target = target.next) {
+        target.val = 0;
+    }
+    return dummy.val ? dummy : dummy.next;
+}
+```
+
 <!-- tabs:end -->
 
 <!-- solution:end -->

solution/0300-0399/0369.Plus One Linked List/Solution.cpp

@@ -13,16 +13,15 @@ class Solution {
     ListNode* plusOne(ListNode* head) {
         ListNode* dummy = new ListNode(0, head);
         ListNode* target = dummy;
-        while (head) {
-            if (head->val != 9) target = head;
-            head = head->next;
+        for (; head; head = head->next) {
+            if (head->val != 9) {
+                target = head;
+            }
         }
-        ++target->val;
-        target = target->next;
-        while (target) {
+        target->val++;
+        for (target = target->next; target; target = target->next) {
             target->val = 0;
-            target = target->next;
         }
-        return dummy->val == 1 ? dummy : dummy->next;
+        return dummy->val ? dummy : dummy->next;
     }
 };

solution/0300-0399/0369.Plus One Linked List/Solution.go

@@ -15,10 +15,8 @@ func plusOne(head *ListNode) *ListNode {
 		head = head.Next
 	}
 	target.Val++
-	target = target.Next
-	for target != nil {
+	for target = target.Next; target != nil; target = target.Next {
 		target.Val = 0
-		target = target.Next
 	}
 	if dummy.Val == 1 {
 		return dummy

solution/0300-0399/0369.Plus One Linked List/Solution.py

@@ -4,7 +4,7 @@
 #         self.val = val
 #         self.next = next
 class Solution:
-    def plusOne(self, head: ListNode) -> ListNode:
+    def plusOne(self, head: Optional[ListNode]) -> Optional[ListNode]:
         dummy = ListNode(0, head)
         target = dummy
         while head:

solution/0300-0399/0369.Plus One Linked List/Solution.ts

@@ -0,0 +1,27 @@
+/**
+ * Definition for singly-linked list.
+ * class ListNode {
+ *     val: number
+ *     next: ListNode | null
+ *     constructor(val?: number, next?: ListNode | null) {
+ *         this.val = (val===undefined ? 0 : val)
+ *         this.next = (next===undefined ? null : next)
+ *     }
+ * }
+ */
+
+function plusOne(head: ListNode | null): ListNode | null {
+    const dummy = new ListNode(0, head);
+    let target = dummy;
+    while (head) {
+        if (head.val !== 9) {
+            target = head;
+        }
+        head = head.next;
+    }
+    target.val++;
+    for (target = target.next; target; target = target.next) {
+        target.val = 0;
+    }
+    return dummy.val ? dummy : dummy.next;
+}

solution/0300-0399/0387.First Unique Character in a String/README.md

@@ -59,15 +59,13 @@ tags:
 
 <!-- solution:start -->
 
-### 方法一：数组或哈希表
+### 方法一：计数
 
-我们可以用数组或哈希表 $cnt$ 记录字符串 $s$ 中每个字符出现的次数。
+我们用一个哈希表或者一个长度为 $26$ 的数组 $\text{cnt}$ 来存储每个字符出现的次数，然后从头开始遍历每个字符 $\text{s[i]}$，如果 $\text{cnt[s[i]]}$ 为 $1$，则返回 $i$。
 
-然后我们再遍历字符串 $s$，当遍历到某个字符 $c$ 时，如果 $cnt[c]=1$，则说明 $c$ 是第一个不重复的字符，返回它的索引即可。
+遍历结束后，如果没有找到符合条件的字符，返回 $-1$。
 
-如果遍历完字符串 $s$ 仍然没有找到不重复的字符，返回 $-1$。
-
-时间复杂度 $O(n)$，空间复杂度 $O(\Sigma)$，其中 $\Sigma$ 是字符集的大小。
+时间复杂度 $O(n)$，其中 $n$ 是字符串的长度。空间复杂度 $O(|\Sigma|)$，其中 $\Sigma$ 是字符集，本题中字符集为小写字母，所以 $|\Sigma|=26$。
 
 <!-- tabs:start -->
 
@@ -145,12 +143,12 @@ func firstUniqChar(s string) int {
 
 ```ts
 function firstUniqChar(s: string): number {
-    const cnt = new Array(26).fill(0);
+    const cnt = new Map<string, number>();
     for (const c of s) {
-        cnt[c.charCodeAt(0) - 97]++;
+        cnt.set(c, (cnt.get(c) || 0) + 1);
     }
-    for (let i = 0; i < s.length; i++) {
-        if (cnt[s.charCodeAt(i) - 97] === 1) {
+    for (let i = 0; i < s.length; ++i) {
+        if (cnt.get(s[i]) === 1) {
             return i;
         }
     }
@@ -166,12 +164,12 @@ function firstUniqChar(s: string): number {
  * @return {number}
  */
 var firstUniqChar = function (s) {
-    const cnt = new Array(26).fill(0);
+    const cnt = new Map();
     for (const c of s) {
-        ++cnt[c.charCodeAt() - 'a'.charCodeAt()];
+        cnt.set(c, (cnt.get(c) || 0) + 1);
     }
     for (let i = 0; i < s.length; ++i) {
-        if (cnt[s[i].charCodeAt() - 'a'.charCodeAt()] === 1) {
+        if (cnt.get(s[i]) === 1) {
             return i;
         }
     }

solution/0300-0399/0387.First Unique Character in a String/README_EN.md

@@ -64,7 +64,13 @@ tags:
 
 <!-- solution:start -->
 
-### Solution 1
+### Solution 1: Counting
+
+We use a hash table or an array of length $26$ $\text{cnt}$ to store the frequency of each character. Then, we traverse each character $\text{s[i]}$ from the beginning. If $\text{cnt[s[i]]}$ is $1$, we return $i$.
+
+If no such character is found after the traversal, we return $-1$.
+
+The time complexity is $O(n)$, where $n$ is the length of the string. The space complexity is $O(|\Sigma|)$, where $\Sigma$ is the character set. In this problem, the character set consists of lowercase letters, so $|\Sigma|=26$.
 
 <!-- tabs:start -->
 
@@ -142,12 +148,12 @@ func firstUniqChar(s string) int {
 
 ```ts
 function firstUniqChar(s: string): number {
-    const cnt = new Array(26).fill(0);
+    const cnt = new Map<string, number>();
     for (const c of s) {
-        cnt[c.charCodeAt(0) - 97]++;
+        cnt.set(c, (cnt.get(c) || 0) + 1);
     }
-    for (let i = 0; i < s.length; i++) {
-        if (cnt[s.charCodeAt(i) - 97] === 1) {
+    for (let i = 0; i < s.length; ++i) {
+        if (cnt.get(s[i]) === 1) {
             return i;
         }
     }
@@ -163,12 +169,12 @@ function firstUniqChar(s: string): number {
  * @return {number}
  */
 var firstUniqChar = function (s) {
-    const cnt = new Array(26).fill(0);
+    const cnt = new Map();
     for (const c of s) {
-        ++cnt[c.charCodeAt() - 'a'.charCodeAt()];
+        cnt.set(c, (cnt.get(c) || 0) + 1);
     }
     for (let i = 0; i < s.length; ++i) {
-        if (cnt[s[i].charCodeAt() - 'a'.charCodeAt()] === 1) {
+        if (cnt.get(s[i]) === 1) {
             return i;
         }
     }

solution/0300-0399/0387.First Unique Character in a String/Solution.js

@@ -3,12 +3,12 @@
  * @return {number}
  */
 var firstUniqChar = function (s) {
-    const cnt = new Array(26).fill(0);
+    const cnt = new Map();
     for (const c of s) {
-        ++cnt[c.charCodeAt() - 'a'.charCodeAt()];
+        cnt.set(c, (cnt.get(c) || 0) + 1);
     }
     for (let i = 0; i < s.length; ++i) {
-        if (cnt[s[i].charCodeAt() - 'a'.charCodeAt()] === 1) {
+        if (cnt.get(s[i]) === 1) {
             return i;
         }
     }