常见的一道笔试题：链表拆分，将一个单向链表和拆分成两个，如下所示：
原始链表：1—>2—>3—>4—>5—>6—>7
拆分后链表A：1—>3—>5—>7
拆分后链表B：2—>4—>6
思路：
定义两个头节点，作为新链表的头，分别指向原链表的前两个元素
再定义两个浮动，初始值分别指向新链表的头结点
然后交替遍历原始节点，将各自的next赋值给新链表
待遍历到最后，需要为两个新链表指定尾节点，即赋值为null
最后退出循环。
class ShowMeBug {
public static class Node {
public int data;
public Node next;
public Node(int data, Node nxt) {
this.data = data;
this.next = nxt;
}
}
/**
* 思路，定义两个浮动游标，随之链表移动而移动，完成整个链表的遍历
* @param head
*/
public void solution(Node head) {
Node first = head;
Node second = head.next;
Node curseA = first;//游标A
Node curseB = second;//游标B
while (curseB != null) {
if (curseB.next != null) {
curseA.next = curseB.next;
curseA = curseA.next;
curseB.next = curseA.next;
curseB = curseB.next;
} else {
curseA.next = null;
curseB.next = null;
break;
}
}
printNode(first);
printNode(second);
}
private static void printNode(Node head) {
while (head != null) {
System.out.println(head.data);
head = head.next;
}
}
public static void main(String[] args) {
ShowMeBug showMeBug = new ShowMeBug();
Node node7 = new Node(7, null);
Node node6 = new Node(6, node7);
Node node5 = new Node(5, node6);
Node node4 = new Node(4, node5);
Node node3 = new Node(3, node4);
Node node2 = new Node(2, node3);
Node node1 = new Node(1, node2);
showMeBug.solution(node1);
//printNode(node1);
}
}
