feat: add solutions to lcof problem: No.18

Author: yanglbme

lcof/面试题18. 删除链表的节点/README.md

@@ -33,9 +33,15 @@
 
 ## 解法
 
-定义一个虚拟头节点 `dummy` 指向 `head`，`pre` 指针初始指向 `dummy`。
+**方法一：模拟**
 
-循环遍历链表，`pre` 往后移动。当指针 `pre.next` 指向的节点的值等于 `val` 时退出循环，将 `pre.next` 指向 `pre.next.next`，然后返回 `dummy.next`。
+我们先创建一个虚拟头节点 `dummy`，令 `dummy.next = head`，然后创建一个指针 `cur` 指向 `dummy`。
+
+遍历链表，当 `cur.next.val == val` 时，将 `cur.next` 指向 `cur.next.next`，然后跳出循环。
+
+最后返回 `dummy.next` 即可。
+
+时间复杂度 $O(n)$，空间复杂度 $O(1)$。其中 $n$ 为链表的长度。
 
 <!-- tabs:start -->
 
@@ -49,10 +55,12 @@
 #         self.next = None
 class Solution:
     def deleteNode(self, head: ListNode, val: int) -> ListNode:
-        pre = dummy = ListNode(next=head)
-        while pre.next and pre.next.val != val:
-            pre = pre.next
-        pre.next = None if not pre.next else pre.next.next
+        dummy = cur = ListNode(0, head)
+        while cur.next:
+            if cur.next.val == val:
+                cur.next = cur.next.next
+                break
+            cur = cur.next
         return dummy.next
 ```
 
@@ -68,39 +76,42 @@ class Solution:
  * }
  */
 class Solution {
-
     public ListNode deleteNode(ListNode head, int val) {
         ListNode dummy = new ListNode(0, head);
-        ListNode pre = dummy;
-        for (; pre.next != null && pre.next.val != val; pre = pre.next)
-            ;
-        pre.next = pre.next == null ? null : pre.next.next;
+        for (ListNode cur = dummy; cur.next != null; cur = cur.next) {
+            if (cur.next.val == val) {
+                cur.next = cur.next.next;
+                break;
+            }
+        }
         return dummy.next;
     }
 }
 ```
 
-### **JavaScript**
+### **C++**
 
-```js
+```cpp
 /**
  * Definition for singly-linked list.
- * function ListNode(val) {
- *     this.val = val;
- *     this.next = null;
- * }
- */
-/**
- * @param {ListNode} head
- * @param {number} val
- * @return {ListNode}
+ * struct ListNode {
+ *     int val;
+ *     ListNode *next;
+ *     ListNode(int x) : val(x), next(NULL) {}
+ * };
  */
-var deleteNode = function (head, val) {
-    const dummy = new ListNode(0, head);
-    let pre = dummy;
-    for (; pre.next && pre.next.val != val; pre = pre.next);
-    pre.next = pre.next?.next;
-    return dummy.next;
+class Solution {
+public:
+    ListNode* deleteNode(ListNode* head, int val) {
+        ListNode* dummy = new ListNode(0, head);
+        for (ListNode* cur = dummy; cur->next; cur = cur->next) {
+            if (cur->next->val == val) {
+                cur->next = cur->next->next;
+                break;
+            }
+        }
+        return dummy->next;
+    }
 };
 ```
 
@@ -116,37 +127,40 @@ var deleteNode = function (head, val) {
  */
 func deleteNode(head *ListNode, val int) *ListNode {
 	dummy := &ListNode{0, head}
-	pre := dummy
-	for ; pre.Next != nil && pre.Next.Val != val; pre = pre.Next {
-	}
-	if pre.Next != nil {
-		pre.Next = pre.Next.Next
+	for cur := dummy; cur.Next != nil; cur = cur.Next {
+		if cur.Next.Val == val {
+			cur.Next = cur.Next.Next
+			break
+		}
 	}
 	return dummy.Next
 }
 ```
 
-### **C++**
+### **JavaScript**
 
-```cpp
+```js
 /**
  * Definition for singly-linked list.
- * struct ListNode {
- *     int val;
- *     ListNode *next;
- *     ListNode(int x) : val(x), next(NULL) {}
- * };
+ * function ListNode(val) {
+ *     this.val = val;
+ *     this.next = null;
+ * }
  */
-class Solution {
-public:
-    ListNode* deleteNode(ListNode* head, int val) {
-        ListNode* dummy = new ListNode(0, head);
-        ListNode* pre = dummy;
-        for (; pre->next && pre->next->val != val; pre = pre->next)
-            ;
-        pre->next = pre->next ? pre->next->next : nullptr;
-        return dummy->next;
+/**
+ * @param {ListNode} head
+ * @param {number} val
+ * @return {ListNode}
+ */
+var deleteNode = function (head, val) {
+    const dummy = new ListNode(0, head);
+    for (let cur = dummy; cur.next; cur = cur.next) {
+        if (cur.next.val == val) {
+            cur.next = cur.next.next;
+            break;
+        }
     }
+    return dummy.next;
 };
 ```
 
@@ -195,23 +209,18 @@ impl Solution {
  *     public ListNode(int x) { val = x; }
  * }
  */
-
- public class Solution {
-     public ListNode DeleteNode(ListNode head, int val) {
-         if (head == null) {
-             return null;
-         }
-         if (head.val == val) {
-             return head.next;
-         }
-         ListNode p = head;
-         while (p.next != null && p.next.val != val) {
-             p = p.next;
-         }
-         p.next = p.next == null ? null : p.next.next;
-         return head;
-     }
- }
+public class Solution {
+    public ListNode DeleteNode(ListNode head, int val) {
+        ListNode dummy = new ListNode(0, head);
+        for (ListNode cur = dummy; cur.next != null; cur = cur.next) {
+            if (cur.next.val == val) {
+                cur.next = cur.next.next;
+                break;
+            }
+        }
+        return dummy.next;
+    }
+}
 ```
 
 ### **...**

lcof/面试题18. 删除链表的节点/Solution.cpp

@@ -10,10 +10,12 @@ class Solution {
 public:
     ListNode* deleteNode(ListNode* head, int val) {
         ListNode* dummy = new ListNode(0, head);
-        ListNode* pre = dummy;
-        for (; pre->next && pre->next->val != val; pre = pre->next)
-            ;
-        pre->next = pre->next ? pre->next->next : nullptr;
+        for (ListNode* cur = dummy; cur->next; cur = cur->next) {
+            if (cur->next->val == val) {
+                cur->next = cur->next->next;
+                break;
+            }
+        }
         return dummy->next;
     }
 };

lcof/面试题18. 删除链表的节点/Solution.cs

@@ -6,20 +6,15 @@
  *     public ListNode(int x) { val = x; }
  * }
  */
-
- public class Solution {
-     public ListNode DeleteNode(ListNode head, int val) {
-         if (head == null) {
-             return null;
-         }
-         if (head.val == val) {
-             return head.next;
-         }
-         ListNode p = head;
-         while (p.next != null && p.next.val != val) {
-             p = p.next;
-         }
-         p.next = p.next == null ? null : p.next.next;
-         return head;
-     }
- }
+public class Solution {
+    public ListNode DeleteNode(ListNode head, int val) {
+        ListNode dummy = new ListNode(0, head);
+        for (ListNode cur = dummy; cur.next != null; cur = cur.next) {
+            if (cur.next.val == val) {
+                cur.next = cur.next.next;
+                break;
+            }
+        }
+        return dummy.next;
+    }
+}

lcof/面试题18. 删除链表的节点/Solution.go

@@ -7,11 +7,11 @@
  */
 func deleteNode(head *ListNode, val int) *ListNode {
 	dummy := &ListNode{0, head}
-	pre := dummy
-	for ; pre.Next != nil && pre.Next.Val != val; pre = pre.Next {
-	}
-	if pre.Next != nil {
-		pre.Next = pre.Next.Next
+	for cur := dummy; cur.Next != nil; cur = cur.Next {
+		if cur.Next.Val == val {
+			cur.Next = cur.Next.Next
+			break
+		}
 	}
 	return dummy.Next
 }

lcof/面试题18. 删除链表的节点/Solution.java

@@ -9,10 +9,12 @@
 class Solution {
     public ListNode deleteNode(ListNode head, int val) {
         ListNode dummy = new ListNode(0, head);
-        ListNode pre = dummy;
-        for (; pre.next != null && pre.next.val != val; pre = pre.next)
-            ;
-        pre.next = pre.next == null ? null : pre.next.next;
+        for (ListNode cur = dummy; cur.next != null; cur = cur.next) {
+            if (cur.next.val == val) {
+                cur.next = cur.next.next;
+                break;
+            }
+        }
         return dummy.next;
     }
 }

lcof/面试题18. 删除链表的节点/Solution.js

@@ -12,8 +12,11 @@
  */
 var deleteNode = function (head, val) {
     const dummy = new ListNode(0, head);
-    let pre = dummy;
-    for (; pre.next && pre.next.val != val; pre = pre.next);
-    pre.next = pre.next?.next;
+    for (let cur = dummy; cur.next; cur = cur.next) {
+        if (cur.next.val == val) {
+            cur.next = cur.next.next;
+            break;
+        }
+    }
     return dummy.next;
 };

lcof/面试题18. 删除链表的节点/Solution.py

@@ -5,8 +5,10 @@
 #         self.next = None
 class Solution:
     def deleteNode(self, head: ListNode, val: int) -> ListNode:
-        pre = dummy = ListNode(next=head)
-        while pre.next and pre.next.val != val:
-            pre = pre.next
-        pre.next = None if not pre.next else pre.next.next
+        dummy = cur = ListNode(0, head)
+        while cur.next:
+            if cur.next.val == val:
+                cur.next = cur.next.next
+                break
+            cur = cur.next
         return dummy.next