剑指：反转链表

题目描述

输入一个链表，反转链表后，输出新链表的表头。

解法

解法一

(a) 为原链表

(b) 反转链表：当i把之前所有的的指针指向前一个结点，导致链表在结点 i、j 之间断裂。

所以，反转时，把 i 指向 h 时，还需一个指针记录 j 结点以防断裂。

public static ListNode reverseList(ListNode head){
        if(head == null || head.next ==null){
            return head;
        }
        
        ListNode p = head;
        ListNode q = null;
        while(p!=null){
            ListNode pnext = p.next; //记录被调整的下一个元素
            if(pnext == null)
                head = p;
            
            p.next = q; //调整指针指向
            
            q = p;
            p = pnext;
        }
        return head;
}

package demo;

public class Solution {
    
    public static class ListNode{
        int val;
        ListNode next = null;
        
        ListNode(int val){
            this.val = val;
        }
    }
    
    public static ListNode reverseList(ListNode head){
        if(head == null || head.next ==null){
            return head;
        }
        
        ListNode p = head;
        ListNode q = null;
        while(p!=null){
            ListNode pnext = p.next; //记录被调整的下一个元素
            if(pnext == null)
                head = p;
            
            p.next = q; //调整指针指向
            
            q = p;
            p = pnext;
        }
        return head;
    }

    public static void main(String[] args) {
        ListNode head = null;//new ListNode(0);
        
        
        ListNode node1 = new ListNode(1); head = node1;
        ListNode node2 = new ListNode(2); node1.next = node2;
        ListNode node3 = new ListNode(3); node2.next = node3;
        ListNode node4 = new ListNode(4); node3.next = node4;
        ListNode node5 = new ListNode(5); node4.next = node5;
        ListNode node6 = new ListNode(6); node5.next = node6;
        ListNode node7 = new ListNode(7); node6.next = node7;
        ListNode node8 = new ListNode(8); node7.next = node8;
        ListNode node9 = new ListNode(9); node8.next = node9;
        ListNode node10 = new ListNode(10); node9.next = node10;
        
        ListNode cur = head;
        while(cur != null){
            System.out.print(cur.val + " ");
            cur = cur.next;
        }
        

        ListNode reshead = reverseList(head);
        System.out.println();
        
        ListNode cur2 = reshead;
        while(cur2 != null){
            System.out.print(cur2.val + " ");
            cur2 = cur2.next;
        }
    }
}

解法二：递归

public static ListNode reverseList(ListNode head){
    if(head == null || head.next ==null){
        return head;
    }
    
    ListNode node = reverseList(head.next);
    ListNode cur = node;
    while(cur.next != null){
        cur = cur.next;  //往回遍历
    }
    cur.next = head;  //指针指向前一个元素
    head.next = null;  //前一个元素指向后的指针设null
    return node;
}