本文共 2953 字,大约阅读时间需要 9 分钟。
给定一个排序链表,删除所有含有重复数字的节点,只保留原始链表中没有重复出现的数字。
举例:
原链表: 1->2->3->3->4->4->5 删除后: 1->2->5解法1:
结合删除重复节点和删除指定节点的方式,可以先删除重复节点,并记录下来重复节点的值,然后再删除指定节点的值即可。
public class Code_06 { public static void main(String[] args) { Code_06 c = new Code_06(); ListNode n = new ListNode(); ListNode head = n; for (int i = 1; i <= 5; i++) { n.next = new ListNode(i); n = n.next; if (i % 2 == 0) { n.next = new ListNode(i); n = n.next; } } System.out.println(head); System.out.println(c.deleteDuplicates(head)); } public ListNode deleteDuplicates(ListNode head) { ListNode cur = head; SetdupSet = new HashSet<>(); while (cur != null && cur.next != null) { if (cur.val != cur.next.val) { cur = cur.next; } else { dupSet.add(cur.val); cur.next = cur.next.next; } } for (Integer val : dupSet) { head = deleteNode(head, val); } return head; } private ListNode deleteNode(ListNode head, int val) { while (head != null) { if (head.val != val) { break; } head = head.next; } ListNode pre = null; ListNode cur = head; while (cur != null) { if (cur.val == val) { pre.next = cur.next; } else { pre = cur; } cur = cur.next; } return head; }}
第一种方式显然不够优雅,换一种思路,通过哑节点+双指针实现
public class Code_06_01 { public static void main(String[] args) { Code_06_01 c = new Code_06_01(); ListNode n = new ListNode(); ListNode head = n; for (int i = 1; i <= 5; i++) { n.next = new ListNode(i); n = n.next; if (i % 2 == 0) { n.next = new ListNode(i); n = n.next; } } System.out.println(head); System.out.println(c.deleteDuplicates(head)); } /** * 哑节点+双指针 ** 构建一个哑节点,让其next指向头位置。 * 再利用双指针,n1,n2,初始都指向head位置,如果n1.next.val==n2.next.val,则让n2向前移动一位,否则n1,n2一起向前移动一位 * * @param head * @return */ public ListNode deleteDuplicates(ListNode head) {
ListNode dummy = new ListNode(); dummy.next = head; ListNode n1 = dummy; ListNode n2 = head; while (n2 != null && n2.next != null) { if (n1.next.val != n2.next.val) { n1 = n1.next; } else { //n2一直移动,直到不等于n1为止 while (n2 != null && n2.next != null && n1.next.val == n2.next.val) { n2 = n2.next; } n1.next = n2.next; } n2 = n2.next; } return dummy.next; }}
转载地址:http://uhlrb.baihongyu.com/