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如何反转一个链表
三个指针，一个指向头节点，另外两个用来指向头节点前后两个节点的位置
// 定义链表节点类
class ListNode {constructor(val) {this.val = val;this.next = null;}
}// 定义反转链表函数
function reverseLinkedList(head) {let prev = null;let current = head;while (current !== null) {const nextTemp = current.next;current.next = prev;prev = current;current = nextTemp;}return prev; // 返回反转后的头节点
}// 示例用法
const head = new ListNode(1);
head.next = new ListNode(2);
head.next.next = new ListNode(3);
head.next.next.next = new ListNode(4);
head.next.next.next.next = new ListNode(5);console.log("原始链表:");
let current = head;
while (current !== null) {console.log(current.val);current = current.next;
}const reversedHead = reverseLinkedList(head);console.log("反转后的链表:");
let reversedCurrent = reversedHead;
while (reversedCurrent !== null) {console.log(reversedCurrent.val);reversedCurrent = reversedCurrent.next;
}
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