链表与递归----力扣203题

力扣算法203题--移除链表元素

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;
    }

    //链表节点的构造函数
    //使用arr为参数，创建一个链表，当前的ListNode为链表头节点
    public ListNode(int[] arr){
        if(arr == null && arr.length == 0)
            throw new IllegalArgumentException("arr cannot be empty.");

        this.val = arr[0];
        ListNode cur = this;
        for (int i = 0; i < arr.length; i++) {
            cur.next = new ListNode(arr[i]);
            cur = cur.next;
        }
    }

    //以当前节点为头节点的链表信息字符串
    @Override
    public String toString(){

        StringBuilder res = new StringBuilder();
        ListNode cur = this;
        while (cur != null){
            res.append(cur.val+"->");
            cur = cur.next;
        }
        res.append("NULL");
        return res.toString();
    }
}

 

解法一：

/**
 * 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; }
 * }
 */
public class Solution {
    public ListNode removeElements(ListNode head, int val) {

        while(head != null && head.val == val){
            ListNode delNode = head;
            head = head.next;
            delNode.next = null;
        }

        if(head == null)
            return head;

        ListNode prev = head;
        while (prev.next != null){
            if(prev.next.val == val){
                ListNode delNode = prev.next;
                prev.next = delNode.next;
                delNode.next = null;
            }else{
                prev = prev.next;
            }
        }

        return head;
    }

    public static void main(String[] args) {
        int[] nums = {1,2,3,6,4,5,6};
        ListNode listNode = new ListNode(nums);
        System.out.println(listNode);

        ListNode res = (new Solution()).removeElements(listNode,6);
        System.out.println(res);
    }
}

运行结果：

1->1->2->3->6->4->5->6->NULL
1->1->2->3->4->5->NULL

Process finished with exit code 0

解法二：使用虚拟头节点

/**
 * 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; }
 * }
 */
//通过虚拟头节点解决问题
public class Solution2 {
    public ListNode removeElements(ListNode head, int val) {

        ListNode dummyHead = new ListNode(-1);
        dummyHead.next = head;

        ListNode prev = dummyHead;
        while (prev.next != null){
            if(prev.next.val == val)
                prev.next = prev.next.next;
            else
                prev = prev.next;
        }

        return dummyHead.next;
    }
}