代码随想录算法训练营第3天 | 链表删除元素、翻转链表

删除链表元素，技巧是设置一个虚拟头节点，这样就可以把原始头节点当做普通节点处理了，最后再返回虚拟头结点的next即可。

题203. 移除链表元素

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode() {}
 *     ListNode(int val) { this.val = val; }
 *     ListNode(int val, ListNode next) { this.val = val; this.next = next; }
 * }
 */
class Solution {
    public ListNode removeElements(ListNode head, int val) {
        if(head==null){
            return null;
        }
        ListNode vir = new ListNode();
        vir.next = head;
        ListNode cur1 = vir;
        ListNode cur2 = vir.next;
        while(cur2!=null){
            if(cur2.val==val){
                cur1.next = cur2.next;
                cur2 = cur1.next;
            }else{
                cur1=cur2;
                cur2=cur2.next;
            }
        }
        return vir.next;
    }
}

反转链表
注意：

• 自己模拟一遍过程，注意最后退出循环后要把新的头节点的next处理好。
• 递归和循环皆可。

//递归
/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode() {}
 *     ListNode(int val) { this.val = val; }
 *     ListNode(int val, ListNode next) { this.val = val; this.next = next; }
 * }
 */
class Solution {
    public ListNode reverseList(ListNode head) {
        if(head==null){
            return null;
        }
        if(head.next==null){
            return head;
        }
        if(head.next.next==null){
            ListNode p1 = head.next;
            p1.next = head;
            head.next = null;
            return p1;
        }
        ListNode cur1 = head;
        ListNode cur2 = head.next;
        ListNode cur3 = head.next.next;
        return reverse(head,cur1,cur2,cur3);
    }

    public ListNode reverse(ListNode head,ListNode cur1,ListNode cur2,ListNode cur3){
        if(cur3==null){
            cur2.next = cur1;
            head.next = null;
            return cur2;
        }else{
            cur2.next = cur1;
            cur1 = cur2;
            cur2 = cur3;
            cur3 = cur3.next;
            return reverse(head,cur1,cur2,cur3);
        }
    }
}

//循环
class Solution {
    public ListNode reverseList(ListNode head) {
        if(head==null){
            return null;
        }
        if(head.next==null){
            return head;
        }
        if(head.next.next==null){
            ListNode p1 = head.next;
            p1.next = head;
            head.next = null;
            return p1;
        }
        ListNode cur1 = head;
        ListNode cur2 = head.next;
        ListNode cur3 = head.next.next;
        do{
            cur2.next = cur1;
            cur1 = cur2;
            cur2 = cur3;
            cur3 = cur3.next;
        }while(cur3!=null);
        cur2.next = cur1;
        head.next = null;
        return cur2;
    }
}