feat: 反转链表(206)

leetcode/简单/反转链表.js

@@ -0,0 +1,106 @@
+/*
+删除排序链表中的重复元素（LeetCode83）206. 反转链表
+简单
+
+相关标签
+相关企业
+给你单链表的头节点 head ，请你反转链表，并返回反转后的链表。
+
+示例 1：
+
+输入：head = [1,2,3,4,5]
+输出：[5,4,3,2,1]
+示例 2：
+
+输入：head = [1,2]
+输出：[2,1]
+示例 3：
+
+输入：head = []
+输出：[]
+
+提示：
+
+链表中节点的数目范围是 [0, 5000]
+-5000 <= Node.val <= 5000
+*/
+
+/**
+ * Definition for singly-linked list.
+ * function ListNode(val, next) {
+ *     this.val = (val===undefined ? 0 : val)
+ *     this.next = (next===undefined ? null : next)
+ * }
+ */
+
+// 思路1
+// const reverseList = function (head) {
+//   if (!head || !head.next) return head; // 如果链表为空或只有一个节点，直接返回
+
+//   const stack = []; // 创建一个栈
+
+//   // 遍历链表，将节点依次压入栈中
+//   let current = head;
+//   while (current !== null) {
+//     stack.push(current);
+//     current = current.next;
+//   }
+
+//   const newHead = stack.pop(); // 取出栈顶节点作为新链表的头节点
+//   let newCurrent = newHead;
+
+//   // 依次弹出栈中的节点，连接成新的链表
+//   while (stack.length > 0) {
+//     newCurrent.next = stack.pop();
+//     newCurrent = newCurrent.next;
+//   }
+
+//   newCurrent.next = null; // 尾节点的 next 指针置空，确保新链表的结束
+
+//   return newHead; // 返回新链表的头节点
+// };
+
+// 思路2 （迭代法）
+
+// const reverseList = function(head) {
+//   let prev = null;
+//   let current = head;
+//   while (current !== null) {
+//       const nextTemp = current.next;
+//       current.next = prev;
+//       prev = current;
+//       current = nextTemp;
+//   }
+//   return prev;
+// };
+
+// 思路3 （递归法）
+import { LinkListNode as ListNode } from '../../list/index.js';
+
+const reverseList = function (head) {
+  if (head == null || head.next === null) return head;
+  const rList = reverseList(head.next);
+  head.next.next = head;
+  head.next = null;
+  return rList;
+};
+// 示例链表：1 -> 2 -> 3 -> 4 -> 5
+const node1 = new ListNode(1);
+const node2 = new ListNode(2);
+const node3 = new ListNode(3);
+const node4 = new ListNode(4);
+const node5 = new ListNode(5);
+node1.next = node2;
+node2.next = node3;
+node3.next = node4;
+node4.next = node5;
+
+// 反转链表
+const reversedHead = reverseList(node1);
+
+// 打印反转后的链表
+let current = reversedHead;
+while (current !== null) {
+  console.log(current.val);
+  current = current.next;
+}