说明:思路中写的是代码,表达基本意思
一、删除链表中所有与val相等的元素
定义两个结点:prev和cur
遍历整个链表:
相等:prve.next=cur.next
cur=cur.next
prev=prev.next
不相等:cur=cur.next
创新互联建站是一家专业提供岱岳企业网站建设,专注与成都网站制作、做网站、H5技术、小程序制作等业务。10年已为岱岳众多企业、政府机构等服务。创新互联专业网络公司优惠进行中。
二、合并两个有序链表
定义两个结点result(合成的新链表的头结点) last(result的最后一个结点)
如果cur1.val<=cur2.val
last.next=cur1
cur1=cur1.next
否则,同理,更新cur2
```class Node {
int val;
Node next = null;
public Node(int val) {
this.val = val;
}
}
public class Solution {
Node removeAll(Node head, int val) {
Node prev = null;
Node cur = head;
while (cur != null) {
if (cur.val == val) {
if (cur == head) {
head = cur.next;
} else {
prev.next = cur.next;
}
} else {
prev = cur;
}
cur = cur.next;
}
return head;
}
Node merge(Node head1, Node head2) {
if (head1 == null) {
return head2;
}
if (head2 == null) {
return head1;
}
Node result = null;
Node last = null;
Node cur1 = head1;
Node cur2 = head2;
while (cur1 != null && cur2 != null) {
if (cur1.val <= cur2.val) {
if (result == null) {
result = cur1;
} else {
last.next = cur1;
}
last = cur1;
cur1 = cur1.next;
} else {
if (result == null) {
result = cur2;
} else {
last.next = cur2;
}
last = cur2;
cur2 = cur2.next;
}
}
if (cur1 != null) {
last.next = cur1;
} else {
last.next = cur2;
}
return result;
}
public static Node createList() {
Node n1 = new Node(6);
Node n3 = new Node(2);
Node n4 = new Node(6);
Node n6 = new Node(4);
Node n8 = new Node(6);
n1.next = n3;
n3.next = n4;
n4.next = n6;
n6.next = n8;
return n1;
}
public static Node createList1() {
Node n1 = new Node(1);
Node n2 = new Node(2);
n1.next = n2;
return n1;
}
public static Node createList2() {
Node n1 = new Node(1);
Node n2 = new Node(3);
Node n3 = new Node(5);
Node n4 = new Node(7);
n1.next = n2;
n2.next = n3;
n3.next = n4;
return n1;
}
public static void main(String[] args) {
Node head = createList();
Node result = new Solution().removeAll(head, 6);
for (Node cur = result; cur != null; cur = cur.next) {
System.out.println(cur.val);
}
System.out.println("=====================");
Node head1 = createList1();
Node head2 = createList2();
Node merged = new Solution().merge(head1, head2);//类中的函数返回值是Node类型,用merge接收
for (Node cur = merged; cur != null; cur = cur.next) {//merge相当于head,代表整个链表
System.out.println(cur.val);
}
}
}