13.在O(1)时间删除链表结点(JAVA)

在O(1)时间删除链表结点

题目描述：

给定单向链表的头指针和一个结点指针，定义一个函数在O(1)时间删除该结点。

具体实现

**解题思路：**要删除这个结点i，可以复制这个结点i的下个结点j的内容到i上，然后把i结点的next指针指向j的下一个结点，再把j结点删除，这样就可以实现删除之前i结点了

public class Shanchu_jiedian {
	static class ListNode{
		int Value;
		ListNode next;
		public ListNode(int Value) {
			// TODO Auto-generated constructor stub
			this.Value = Value;
		}
		public ListNode(int Value, ListNode next) {
			// TODO Auto-generated constructor stub
			this.Value = Value;
			this.next = next;
		}
		public ListNode() {
			// TODO Auto-generated constructor stub
		}
		
	};
	public static void DeleteNode(ListNode headnode, ListNode deletenode) {
		if(headnode == null || deletenode == null)
			return;
		if(deletenode.next != null) {//要删除的结点不是尾结点
			ListNode nextnode = deletenode.next;
			deletenode.Value = nextnode.Value;
			deletenode.next = nextnode.next;
		}else if(headnode == deletenode) {//链表只有一个结点，删除头结点（也是尾结点）
			headnode = null;
			deletenode = null;
		}else {//链表有多个结点，删除结点是尾结点
			ListNode temp = headnode;
			while(temp.next != deletenode) {
				temp = temp.next;
			}
			temp.next = null;
			deletenode = null;
		}
		
	}
	public static void main(String[] args) {
		// TODO Auto-generated method stub
        //node4->node3->node2->node1
		ListNode node1 = new ListNode(11);
		ListNode node2 = new ListNode(2,node1);
		ListNode node3 = new ListNode(3,node2);
		ListNode node4 = new ListNode(4,node3);
		System.out.println(node1 + " " +node1.Value + " " + node1.next);
		System.out.println(node2 + " " +node2.Value + " " + node2.next);
		System.out.println(node3 + " " +node3.Value + " " + node3.next);
		System.out.println(node4 + " " +node4.Value + " " + node4.next);
		System.out.println();
		DeleteNode(node4,node3);
		System.out.println(node1 + " " +node1.Value + " " + node1.next);
		System.out.println(node2 + " " +node2.Value + " " + node2.next);
		System.out.println(node3 + " " +node3.Value + " " + node3.next);
		System.out.println(node4 + " " +node4.Value + " " + node4.next);
	}

}