剑指offer_13(在O(1)的时间内删除链表中的某一个元素)

来源：互联网发布：c语言open函数编辑：程序博客网时间：2024/05/22 11:47

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);    }}

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