剑指offer_13(在O(1)的时间内删除链表中的某一个元素)

题意：在O(1)的时间内删除链表中的某一个元素
思路：若链表只有一个元素，那么就是删除这一唯一元素；若删除的是链表最后一个元素，那么只能从头遍历；若删除的是只是中间的一个元素，那么将待删除的元素的下一个元素的值赋给待删除元素，再将待删除元素的下一个元素给删除即可。
代码：

package MianShiTi_13;

import java.awt.List;

public class MianShiTi_13 {
    public static class ListNode {
        int value;
        ListNode next;
    }
    public static ListNode deleteNode(ListNode head , ListNode toBeDeleted){
        if(head == null || toBeDeleted == null){
            return head;
        }
        if(head == toBeDeleted){
            return head.next;
        }
        if(toBeDeleted.next == null){
            ListNode tmp = head;
            while (tmp.next != toBeDeleted) {
                tmp = tmp.next;
            }
            tmp.next = null;
        }
        else{
            toBeDeleted.value = toBeDeleted.next.value;
            toBeDeleted.next = toBeDeleted.next.next;
        }
        return head;
    }

    public static void printListValue(ListNode head) {
        while (head != null) {
            System.out.print(head.value+" ");
            head = head.next;
        }
    }

    public static void main(String[] args) {
        ListNode head = new ListNode();
        head.value = 1;
        head.next = new ListNode();
        head.next.value = 2;
        ListNode middle = head.next.next = new ListNode();
        head.next.next.value = 3;
        head.next.next.next = new ListNode();
        head.next.next.next.value = 4;
        ListNode last = head.next.next.next.next = new ListNode();
        head.next.next.next.next.value = 5;

        //删除节点为空
        head = deleteNode(head, null);
        printListValue(head);
        System.out.println();

        //删除中间节点
        head = deleteNode(head, middle);
        printListValue(head);   
        System.out.println();

        //删除末尾节点
        head = deleteNode(head, last);
        printListValue(head);
    }
}