【剑指offer】面试题24：反转链表

完整代码地址

完整代码地址

题目

定义一个函数，输入一个链表的头结点，反转该链表并输出反转后链表的头结点

思路

很简单，单纯考察代码的鲁棒性
要针对区分成以下三种情况处理：
1.输入的链表头指针为null
2.输入的链表只有一个节点
3.输入的链表有多个节点（正常情况）

代码

public static class ListNode {    public int val;    public ListNode next = null;    public ListNode(int val) {        this.val = val;    }}public static ListNode ReverseList(ListNode head) {    // 长度为0或1    if(head == null || head.next == null)        return head;    ListNode first = head;    ListNode second = head.next;    head.next = null;    ListNode third;    while(second != null) {        third = second.next;        second.next = first;        first = second;        second = third;    }    return first;}

测试

public static void main(String[] args) {    test1();    test2();    test3();}/** * 功能测试 */private static void test1() {    ListNode node1 = new ListNode(1);    ListNode node2 = new ListNode(2);    ListNode node3 = new ListNode(3);    ListNode node4 = new ListNode(4);    node1.next = node2;    node2.next = node3;    node3.next = node4;    // 4-3-2-1    node1 = _24_ReverseList.ReverseList(node1);    printList(node1);}/** * 边界测试 * 长度为1 */private static void test2() {    ListNode node1 = new ListNode(5);    node1 = _24_ReverseList.ReverseList(node1);    printList(node1);}/** * 极端测试  */private static void test3() {    ListNode node1 = _24_ReverseList.ReverseList(null);    printList(node1);}private static void printList(ListNode head) {    ListNode cur = head;    while(cur != null) {        System.out.print(cur.val + " ");        cur = cur.next;    }    System.out.println();}