doocs · acbin · Apr 2, 2024 · Apr 2, 2024
diff --git a/lcci/02.06.Palindrome Linked List/README.md b/lcci/02.06.Palindrome Linked List/README.md
diff --git a/lcci/02.06.Palindrome Linked List/README_EN.md b/lcci/02.06.Palindrome Linked List/README_EN.md
diff --git a/lcci/02.06.Palindrome Linked List/Solution.cpp b/lcci/02.06.Palindrome Linked List/Solution.cpp
@@ -0,0 +1,40 @@
+/**
+ * Definition for singly-linked list.
+ * struct ListNode {
+ *     int val;
+ *     ListNode *next;
+ *     ListNode(int x) : val(x), next(NULL) {}
+ * };
+ */
+class Solution {
+public:
+    bool isPalindrome(ListNode* head) {
+        if (!head) {
+            return true;
+        }
+        ListNode* slow = head;
+        ListNode* fast = head->next;
+        while (fast && fast->next) {
+            slow = slow->next;
+            fast = fast->next->next;
+        }
+        ListNode* p = slow->next;
+        slow->next = nullptr;
+        ListNode* dummy = new ListNode(0);
+        while (p) {
+            ListNode* next = p->next;
+            p->next = dummy->next;
+            dummy->next = p;
+            p = next;
+        }
+        p = dummy->next;
+        while (p) {
+            if (head->val != p->val) {
+                return false;
+            }
+            head = head->next;
+            p = p->next;
+        }
+        return true;
+    }
+};
diff --git a/lcci/02.06.Palindrome Linked List/Solution.cs b/lcci/02.06.Palindrome Linked List/Solution.cs
@@ -3,44 +3,37 @@
  * public class ListNode {
  *     public int val;
  *     public ListNode next;
- *     public ListNode(int val=0, ListNode next=null) {
- *         this.val = val;
- *         this.next = next;
- *     }
+ *     public ListNode(int x) { val = x; }
  * }
  */
 public class Solution {
     public bool IsPalindrome(ListNode head) {
-        if (head == null || head.next == null)
-        {
+        if (head == null) {
             return true;
         }
         ListNode slow = head;
         ListNode fast = head.next;
-        while (fast != null && fast.next != null)
-        {
+        while (fast != null && fast.next != null) {
             slow = slow.next;
             fast = fast.next.next;
         }
-        ListNode cur = slow.next;
+        ListNode p = slow.next;
         slow.next = null;
-        ListNode pre = null;
-        while (cur != null)
-        {
-            ListNode t = cur.next;
-            cur.next = pre;
-            pre = cur;
-            cur = t;
+        ListNode dummy = new ListNode(0);
+        while (p != null) {
+            ListNode next = p.next;
+            p.next = dummy.next;
+            dummy.next = p;
+            p = next;
         }
-        while (pre != null)
-        {
-            if (pre.val != head.val)
-            {
+        p = dummy.next;
+        while (p != null) {
+            if (head.val != p.val) {
                 return false;
             }
-            pre = pre.next;
             head = head.next;
+            p = p.next;
         }
         return true;
     }
-}
+}
diff --git a/lcci/02.06.Palindrome Linked List/Solution.go b/lcci/02.06.Palindrome Linked List/Solution.go
@@ -1,40 +1,34 @@
+/**
+ * Definition for singly-linked list.
+ * type ListNode struct {
+ *     Val int
+ *     Next *ListNode
+ * }
+ */
 func isPalindrome(head *ListNode) bool {
 	if head == nil {
 		return true
 	}
-	m := mid(head)
-	temp := reverse(m.Next)
-	m.Next = nil
-	p, q := head, temp
-	res := true
-	for p != nil && q != nil {
-		if p.Val != q.Val {
-			res = false
-			break
-		}
-		p = p.Next
-		q = q.Next
-	}
-	m.Next = reverse(temp)
-	return res
-}
-
-func mid(head *ListNode) *ListNode {
 	slow, fast := head, head.Next
 	for fast != nil && fast.Next != nil {
-		slow = slow.Next
-		fast = fast.Next.Next
+		slow, fast = slow.Next, fast.Next.Next
 	}
-	return slow
-}
-
-func reverse(head *ListNode) *ListNode {
-	var prev *ListNode = nil
-	for head != nil {
-		temp := head.Next
-		head.Next = prev
-		prev = head
-		head = temp
+	p := slow.Next
+	slow.Next = nil
+	dummy := &ListNode{}
+	for p != nil {
+		next := p.Next
+		p.Next = dummy.Next
+		dummy.Next = p
+		p = next
+	}
+	p = dummy.Next
+	for p != nil {
+		if head.Val != p.Val {
+			return false
+		}
+		head = head.Next
+		p = p.Next
 	}
-	return prev
+	return true
 }
diff --git a/lcci/02.06.Palindrome Linked List/Solution.java b/lcci/02.06.Palindrome Linked List/Solution.java
@@ -3,14 +3,12 @@
  * public class ListNode {
  *     int val;
  *     ListNode next;
- *     ListNode() {}
- *     ListNode(int val) { this.val = val; }
- *     ListNode(int val, ListNode next) { this.val = val; this.next = next; }
+ *     ListNode(int x) { val = x; }
  * }
  */
 class Solution {
     public boolean isPalindrome(ListNode head) {
-        if (head == null || head.next == null) {
+        if (head == null) {
             return true;
         }
         ListNode slow = head;
@@ -19,21 +17,22 @@ public boolean isPalindrome(ListNode head) {
             slow = slow.next;
             fast = fast.next.next;
         }
-        ListNode cur = slow.next;
+        ListNode p = slow.next;
         slow.next = null;
-        ListNode pre = null;
-        while (cur != null) {
-            ListNode t = cur.next;
-            cur.next = pre;
-            pre = cur;
-            cur = t;
+        ListNode dummy = new ListNode(0);
+        while (p != null) {
+            ListNode next = p.next;
+            p.next = dummy.next;
+            dummy.next = p;
+            p = next;
         }
-        while (pre != null) {
-            if (pre.val != head.val) {
+        p = dummy.next;
+        while (p != null) {
+            if (head.val != p.val) {
                 return false;
             }
-            pre = pre.next;
             head = head.next;
+            p = p.next;
         }
         return true;
     }

diff --git a/lcci/02.06.Palindrome Linked List/Solution.js b/lcci/02.06.Palindrome Linked List/Solution.js
@@ -1,16 +1,16 @@
 /**
  * Definition for singly-linked list.
- * function ListNode(val, next) {
- *     this.val = (val===undefined ? 0 : val)
- *     this.next = (next===undefined ? null : next)
+ * function ListNode(val) {
+ *     this.val = val;
+ *     this.next = null;
  * }
  */
 /**
  * @param {ListNode} head
  * @return {boolean}
  */
 var isPalindrome = function (head) {
-    if (!head || !head.next) {
+    if (!head) {
         return true;
     }
     let slow = head;
@@ -19,21 +19,22 @@ var isPalindrome = function (head) {
         slow = slow.next;
         fast = fast.next.next;
     }
-    let cur = slow.next;
+    let p = slow.next;
     slow.next = null;
-    let pre = null;
-    while (cur) {
-        let t = cur.next;
-        cur.next = pre;
-        pre = cur;
-        cur = t;
+    const dummy = new ListNode(0);
+    while (p) {
+        const next = p.next;
+        p.next = dummy.next;
+        dummy.next = p;
+        p = next;
     }
-    while (pre) {
-        if (pre.val !== head.val) {
+    p = dummy.next;
+    while (p) {
+        if (head.val !== p.val) {
             return false;
         }
-        pre = pre.next;
         head = head.next;
+        p = p.next;
     }
     return true;
 };
diff --git a/lcci/02.06.Palindrome Linked List/Solution.py b/lcci/02.06.Palindrome Linked List/Solution.py
@@ -1,22 +1,28 @@
 # Definition for singly-linked list.
 # class ListNode:
-#     def __init__(self, val=0, next=None):
-#         self.val = val
-#         self.next = next
+#     def __init__(self, x):
+#         self.val = x
+#         self.next = None
 class Solution:
     def isPalindrome(self, head: ListNode) -> bool:
-        if head is None or head.next is None:
+        if head is None:
             return True
         slow, fast = head, head.next
         while fast and fast.next:
-            slow, fast = slow.next, fast.next.next
-        pre, cur = None, slow.next
-        while cur:
-            t = cur.next
-            cur.next = pre
-            pre, cur = cur, t
-        while pre:
-            if pre.val != head.val:
+            slow = slow.next
+            fast = fast.next.next
+        p = slow.next
+        slow.next = None
+        dummy = ListNode()
+        while p:
+            next = p.next
+            p.next = dummy.next
+            dummy.next = p
+            p = next
+        p = dummy.next
+        while p:
+            if head.val != p.val:
                 return False
-            pre, head = pre.next, head.next
+            head = head.next
+            p = p.next
         return True
diff --git a/lcci/02.06.Palindrome Linked List/Solution.ts b/lcci/02.06.Palindrome Linked List/Solution.ts
@@ -11,29 +11,31 @@
  */
 
 function isPalindrome(head: ListNode | null): boolean {
-    if (head == null || head.next == null) return true;
-    // 快慢指针定位到中点
-    let slow: ListNode = head,
-        fast: ListNode = head.next;
-    while (fast != null && fast.next != null) {
+    if (!head) {
+        return true;
+    }
+    let slow = head;
+    let fast = head.next;
+    while (fast && fast.next) {
         slow = slow.next;
         fast = fast.next.next;
     }
-    // 翻转链表
-    let cur: ListNode = slow.next;
+    let p = slow.next;
     slow.next = null;
-    let prev: ListNode = null;
-    while (cur != null) {
-        let t: ListNode = cur.next;
-        cur.next = prev;
-        prev = cur;
-        cur = t;
+    const dummy = new ListNode(0);
+    while (p) {
+        const next = p.next;
+        p.next = dummy.next;
+        dummy.next = p;
+        p = next;
     }
-    // 判断回文
-    while (prev != null) {
-        if (prev.val != head.val) return false;
-        prev = prev.next;
+    p = dummy.next;
+    while (p) {
+        if (head.val !== p.val) {
+            return false;
+        }
         head = head.next;
+        p = p.next;
     }
     return true;
 }
diff --git a/lcci/02.06.Palindrome Linked List/Solution2.ts b/lcci/02.06.Palindrome Linked List/Solution2.ts
diff --git a/lcci/02.08.Linked List Cycle/README.md b/lcci/02.08.Linked List Cycle/README.md
@@ -5,6 +5,7 @@
 ## 题目描述
 
 <!-- 这里写题目描述 -->
+
 <p>给定一个有环链表，实现一个算法返回环路的开头节点。<br>有环链表的定义：在链表中某个节点的next元素指向在它前面出现过的节点，则表明该链表存在环路。</p><br><p><strong>示例 1：</strong><pre><strong>输入：</strong>head = [3,2,0,-4], pos = 1<br><strong>输出：</strong>tail connects to node index 1<br><strong>解释：</strong>链表中有一个环，其尾部连接到第二个节点。</pre></p><br><p><strong>示例 2：</strong><pre><strong>输入：</strong>head = [1,2], pos = 0<br><strong>输出：</strong>tail connects to node index 0<br><strong>解释：</strong>链表中有一个环，其尾部连接到第一个节点。</pre></p><br><p><strong>示例 3：</strong><pre><strong>输入：</strong>head = [1], pos = -1<br><strong>输出：</strong>no cycle<br><strong>解释：</strong>链表中没有环。</pre></p><br><p><strong>进阶：</strong><br>你是否可以不用额外空间解决此题？</p>
 
 ## 解法