剑指offer 15：反转链表

佚名 6年前 (2019-04-21) 随笔 2146人围观抢沙发百度已收录

题目描述

输入一个链表，反转链表后，输出新链表的表头。

法一：迭代法

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/*
public class ListNode {
    int val;
    ListNode next = null;

    ListNode(int val) {
        this.val = val;
    }
}*/
public class Solution {
    public ListNode ReverseList(ListNode head) {
        if(head == null)
            return null;
        ListNode pre = null;
        ListNode next = null;
      
        while(head != null){
            next = head.next;
            head.next = pre;
            pre = head;
            head = next;
        }
        return pre;
    }
}

法二：递归法

/*
public class ListNode {
    int val;
    ListNode next = null;

    ListNode(int val) {
        this.val = val;
    }
}*/
public class Solution {
    public ListNode ReverseList(ListNode head) {
        if(head == null ||head.next == null)
            return head;
        ListNode next = head.next;
        head.next = null;
        ListNode newHead = ReverseList(next);
        next.next = head;
        return newHead;
    }
}