题目描述:
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。
代码:
public class printCommonPart { public static void printCommonPart(Node head1,Node head2) { System.out.println("Common Part:"); while (head1!=null&&head2!=null) { if(head1.value测试代码:
public class test { public static void main(String[] args) { printCommonPart test = new printCommonPart(); Node head1 = new Node(0); Node head2 = new Node(2); Node cur = head1; for(int i = 1;i<10;i++) { Node temp = new Node(i); cur.next = temp; cur = temp; } Node cur1 = head2; for(int i = 3;i<15;i++) { Node temp = new Node(i); cur1.next = temp; cur1 = temp; } test.printCommonPart(head1,head2); } }结果输出:
说明:
前一段时间一直在刷LeetCode的每日一题,感觉知识点太琐碎。于是准备按题型开刷,找到了左神的书,写的非常好。每天学一点,加油!欢迎分享,转载请注明来源:内存溢出
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