打印两个有序链表的公共部分。
题目
给定两个有序链表的头指针head1和head2，打印两个链表的公共部分。
解答
核心思路
因为两个链表是有序的，所以把两个链表的节点挨个比对，如果比另外一个节点的值小，这个节点就跳到下一个节点，再进行比对，如果相同的话就输出。当其中一条链表遍历完成的时候就结束。
完整带测试代码
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| public class Solution {
public static class Node {
public int value;
public Node next;
public Node(int data) {
this.value = data;
}
}
public static void printCommonPart(Node head1, Node head2) {
System.out.print("Common Part: ");
while (head1 != null && head2 != null) {
if (head1.value < head2.value) {
head1 = head1.next;
} else if (head1.value > head2.value) {
head2 = head2.next;
} else {
System.out.print(head1.value + " ");
head1 = head1.next;
head2 = head2.next;
}
}
System.out.println();
}
public static void printLinkedList(Node node) {
System.out.print("Linked List: ");
while (node != null) {
System.out.print(node.value + " ");
node = node.next;
}
System.out.println();
}
public static void main(String[] args) {
Node node1 = new Node(2);
node1.next = new Node(3);
node1.next.next = new Node(5);
node1.next.next.next = new Node(6);
Node node2 = new Node(1);
node2.next = new Node(2);
node2.next.next = new Node(5);
node2.next.next.next = new Node(7);
node2.next.next.next.next = new Node(8);
printLinkedList(node1);
printLinkedList(node2);
printCommonPart(node1, node2);
}
}
|