package java_jianzhioffer_algorithm;
/**
* 题目:在o(1)内删除链表节点。给定单向链表的头指针和一个节点指针 。
* 假设:删除的节点在链表中
* @author hexiaoli
* 思路:不用遍历找这个节点的前一个节点,而是将这个节点后面节点的值赋给它,再删除这个节点。
* 这个节点中只有一个节点(尾节点),需要将头结点置为空;
* 删除的是尾节点,只能是常规操作.
*/
class Node_l{
Node_l next = null;
int data;
public Node_l(int data ) {
this.data = data;
}
}
public class DeleteListNode {
public static void deleteNode(Node_l head,Node_l toBeDetele) {
//判断输入是否合法,链表是否为空,需要删除的节点是否为空
if(head == null || toBeDetele == null ) {
return; }
//删除的是头节点
if(head == toBeDetele) {
head= null;
}//删除的是尾节点
else if(toBeDetele.next == null) {
while(head.next != toBeDetele) {
head = head.next;
}
head.next = null;
}else {//删除的是中间节点
toBeDetele.data = toBeDetele.next.data;
toBeDetele.next = toBeDetele.next.next;
}
}
public static void main(String[] args) {
//输入链表为空
Node_l head2 = null;
Node_l toBeDetele2;
toBeDetele2 = head2;
deleteNode(head2,toBeDetele2);
//删除的是头节点
Node_l head1 = new Node_l(5);
Node_l toBeDetele1;
toBeDetele1 = head1;
deleteNode(head1,toBeDetele1);
//删除的是尾节点
Node_l head3 = new Node_l(5);
Node_l Node31 = new Node_l(1);
Node_l Node32 = new Node_l(2);
Node_l tail3 = new Node_l(3);
head3.next = Node31;
Node31.next=Node32;
Node32.next=tail3;
Node_l toBeDetele3;
toBeDetele3 = tail3;
deleteNode(head3,toBeDetele3);
//删除的是中间节点
Node_l head4 = new Node_l(5);
Node_l Node41 = new Node_l(1);
Node_l Node42 = new Node_l(2);
Node_l tail4 = new Node_l(3);
head4.next = Node41;
Node41.next=Node42;
Node42.next=tail4;
Node_l toBeDetele4;
toBeDetele4 = Node42;
deleteNode(head4,toBeDetele4);
System.out.println(Node41.next.data);
}
}
剑指offer书(18)删除链表节点
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转载自blog.csdn.net/hxl0925/article/details/89453644
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