feat: update solutions to lc problems: No.83,95 (#2167)

Author: yanglbme

solution/0000-0099/0083.Remove Duplicates from Sorted List/README.md

@@ -38,6 +38,14 @@
 
 <!-- 这里可写通用的实现逻辑 -->
 
+**方法一：一次遍历**
+
+我们用一个指针 $cur$ 来遍历链表。如果当前 $cur$ 与 $cur.next$ 对应的元素相同，我们就将 $cur$ 的 $next$ 指针指向 $cur$ 的下下个节点。否则，说明链表中 $cur$ 对应的元素是不重复的，因此可以将 $cur$ 指针移动到下一个节点。
+
+遍历结束后，返回链表的头节点即可。
+
+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 -->
 
 ### **Python3**
@@ -61,25 +69,6 @@ class Solution:
         return head
 ```
 
-```python
-# Definition for singly-linked list.
-# class ListNode:
-#     def __init__(self, val=0, next=None):
-#         self.val = val
-#         self.next = next
-class Solution:
-    def deleteDuplicates(self, head: Optional[ListNode]) -> Optional[ListNode]:
-        dummy = ListNode(1000)
-        cur = dummy
-        while head:
-            if head.val != cur.val:
-                cur.next = head
-                cur = cur.next
-            head = head.next
-        cur.next = None
-        return dummy.next
-```
-
 ### **Java**
 
 <!-- 这里可写当前语言的特殊实现逻辑 -->
@@ -142,19 +131,53 @@ public:
 ### **Go**
 
 ```go
+/**
+ * Definition for singly-linked list.
+ * type ListNode struct {
+ *     Val int
+ *     Next *ListNode
+ * }
+ */
 func deleteDuplicates(head *ListNode) *ListNode {
-	current := head
-	for current != nil && current.Next != nil {
-		if current.Val == current.Next.Val {
-			current.Next = current.Next.Next
+	cur := head
+	for cur != nil && cur.Next != nil {
+		if cur.Val == cur.Next.Val {
+			cur.Next = cur.Next.Next
 		} else {
-			current = current.Next
+			cur = cur.Next
 		}
 	}
 	return head
 }
 ```
 
+### **JavaScript**
+
+```js
+/**
+ * Definition for singly-linked list.
+ * function ListNode(val) {
+ *     this.val = val;
+ *     this.next = null;
+ * }
+ */
+/**
+ * @param {ListNode} head
+ * @return {ListNode}
+ */
+var deleteDuplicates = function (head) {
+    let cur = head;
+    while (cur && cur.next) {
+        if (cur.next.val === cur.val) {
+            cur.next = cur.next.next;
+        } else {
+            cur = cur.next;
+        }
+    }
+    return head;
+};
+```
+
 ### **Rust**
 
 ```rust
@@ -192,6 +215,35 @@ impl Solution {
 }
 ```
 
+### **C#**
+
+```cs
+/**
+ * Definition for singly-linked list.
+ * public class ListNode {
+ *     public int val;
+ *     public ListNode next;
+ *     public ListNode(int val=0, ListNode next=null) {
+ *         this.val = val;
+ *         this.next = next;
+ *     }
+ * }
+ */
+public class Solution {
+    public ListNode DeleteDuplicates(ListNode head) {
+        ListNode cur = head;
+        while (cur != null && cur.next != null) {
+            if (cur.val == cur.next.val) {
+                cur.next = cur.next.next;
+            } else {
+                cur = cur.next;
+            }
+        }
+        return head;
+    }
+}
+```
+
 ### **...**
 
 ```

solution/0000-0099/0083.Remove Duplicates from Sorted List/README_EN.md

@@ -32,6 +32,14 @@
 
 ## Solutions
 
+**Solution 1: Single Pass**
+
+We use a pointer $cur$ to traverse the linked list. If the element corresponding to the current $cur$ is the same as the element corresponding to $cur.next$, we set the $next$ pointer of $cur$ to point to the next node of $cur.next$. Otherwise, it means that the element corresponding to $cur$ in the linked list is not duplicated, so we can move the $cur$ pointer to the next node.
+
+After the traversal ends, return the head node of the linked list.
+
+The time complexity is $O(n)$, where $n$ is the length of the linked list. The space complexity is $O(1)$.
+
 <!-- tabs:start -->
 
 ### **Python3**
@@ -53,25 +61,6 @@ class Solution:
         return head
 ```
 
-```python
-# Definition for singly-linked list.
-# class ListNode:
-#     def __init__(self, val=0, next=None):
-#         self.val = val
-#         self.next = next
-class Solution:
-    def deleteDuplicates(self, head: Optional[ListNode]) -> Optional[ListNode]:
-        dummy = ListNode(1000)
-        cur = dummy
-        while head:
-            if head.val != cur.val:
-                cur.next = head
-                cur = cur.next
-            head = head.next
-        cur.next = None
-        return dummy.next
-```
-
 ### **Java**
 
 ```java
@@ -132,19 +121,53 @@ public:
 ### **Go**
 
 ```go
+/**
+ * Definition for singly-linked list.
+ * type ListNode struct {
+ *     Val int
+ *     Next *ListNode
+ * }
+ */
 func deleteDuplicates(head *ListNode) *ListNode {
-	current := head
-	for current != nil && current.Next != nil {
-		if current.Val == current.Next.Val {
-			current.Next = current.Next.Next
+	cur := head
+	for cur != nil && cur.Next != nil {
+		if cur.Val == cur.Next.Val {
+			cur.Next = cur.Next.Next
 		} else {
-			current = current.Next
+			cur = cur.Next
 		}
 	}
 	return head
 }
 ```
 
+### **JavaScript**
+
+```js
+/**
+ * Definition for singly-linked list.
+ * function ListNode(val) {
+ *     this.val = val;
+ *     this.next = null;
+ * }
+ */
+/**
+ * @param {ListNode} head
+ * @return {ListNode}
+ */
+var deleteDuplicates = function (head) {
+    let cur = head;
+    while (cur && cur.next) {
+        if (cur.next.val === cur.val) {
+            cur.next = cur.next.next;
+        } else {
+            cur = cur.next;
+        }
+    }
+    return head;
+};
+```
+
 ### **Rust**
 
 ```rust
@@ -182,6 +205,35 @@ impl Solution {
 }
 ```
 
+### **C#**
+
+```cs
+/**
+ * Definition for singly-linked list.
+ * public class ListNode {
+ *     public int val;
+ *     public ListNode next;
+ *     public ListNode(int val=0, ListNode next=null) {
+ *         this.val = val;
+ *         this.next = next;
+ *     }
+ * }
+ */
+public class Solution {
+    public ListNode DeleteDuplicates(ListNode head) {
+        ListNode cur = head;
+        while (cur != null && cur.next != null) {
+            if (cur.val == cur.next.val) {
+                cur.next = cur.next.next;
+            } else {
+                cur = cur.next;
+            }
+        }
+        return head;
+    }
+}
+```
+
 ### **...**
 
 ```

solution/0000-0099/0083.Remove Duplicates from Sorted List/Solution.cs

@@ -1,21 +1,24 @@
-public class Solution {
-    public ListNode DeleteDuplicates(ListNode head) {
-        if (head == null) return null;
-        var last = head;
-        var current = head.next;
-        while (current != null)
-        {
-            if (current.val != last.val)
-            {
-                last.next = current;
-                last = current;
-            }
-            else
-            {
-                last.next = null;
-            }
-            current = current.next;
-        }
-        return head;
-    }
+/**
+ * Definition for singly-linked list.
+ * public class ListNode {
+ *     public int val;
+ *     public ListNode next;
+ *     public ListNode(int val=0, ListNode next=null) {
+ *         this.val = val;
+ *         this.next = next;
+ *     }
+ * }
+ */
+public class Solution {
+    public ListNode DeleteDuplicates(ListNode head) {
+        ListNode cur = head;
+        while (cur != null && cur.next != null) {
+            if (cur.val == cur.next.val) {
+                cur.next = cur.next.next;
+            } else {
+                cur = cur.next;
+            }
+        }
+        return head;
+    }
 }

solution/0000-0099/0083.Remove Duplicates from Sorted List/Solution.go

@@ -1,10 +1,17 @@
+/**
+ * Definition for singly-linked list.
+ * type ListNode struct {
+ *     Val int
+ *     Next *ListNode
+ * }
+ */
 func deleteDuplicates(head *ListNode) *ListNode {
-	current := head
-	for current != nil && current.Next != nil {
-		if current.Val == current.Next.Val {
-			current.Next = current.Next.Next
+	cur := head
+	for cur != nil && cur.Next != nil {
+		if cur.Val == cur.Next.Val {
+			cur.Next = cur.Next.Next
 		} else {
-			current = current.Next
+			cur = cur.Next
 		}
 	}
 	return head

solution/0000-0099/0083.Remove Duplicates from Sorted List/Solution.js

@@ -10,12 +10,12 @@
  * @return {ListNode}
  */
 var deleteDuplicates = function (head) {
-    var p = head;
-    while (p && p.next) {
-        if (p.val == p.next.val) {
-            p.next = p.next.next;
+    let cur = head;
+    while (cur && cur.next) {
+        if (cur.next.val === cur.val) {
+            cur.next = cur.next.next;
         } else {
-            p = p.next;
+            cur = cur.next;
         }
     }
     return head;

solution/0000-0099/0083.Remove Duplicates from Sorted List/Solution.py

@@ -1,14 +1,14 @@
-# Definition for singly-linked list.
-# class ListNode:
-#     def __init__(self, val=0, next=None):
-#         self.val = val
-#         self.next = next
-class Solution:
-    def deleteDuplicates(self, head: ListNode) -> ListNode:
-        cur = head
-        while cur and cur.next:
-            if cur.val == cur.next.val:
-                cur.next = cur.next.next
-            else:
-                cur = cur.next
-        return head
+# Definition for singly-linked list.
+# class ListNode:
+#     def __init__(self, val=0, next=None):
+#         self.val = val
+#         self.next = next
+class Solution:
+    def deleteDuplicates(self, head: Optional[ListNode]) -> Optional[ListNode]:
+        cur = head
+        while cur and cur.next:
+            if cur.val == cur.next.val:
+                cur.next = cur.next.next
+            else:
+                cur = cur.next
+        return head