删除排序链表中的重复元素 II
题目介绍
存在一个按升序排列的链表，给你这个链表的头节点 head ，请你删除链表中所有存在数字重复情况的节点，只保留原始链表中 没有重复出现 的数字。
返回同样按升序排列的结果链表。
示例1：
示例 2：
提示：
- 链表中节点数目在范围
[0, 300] 内
-100 <= Node.val <= 100- 题目数据保证链表已经按升序排列
题目解法
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| package algorithm;
public class RemoveDuplicatesFromSortedListII {
public static ListNode deleteDuplicates(ListNode head) {
if (head == null) {
return head;
}
ListNode dummy = new ListNode(0, head);
ListNode cur = dummy;
while (cur.next != null && cur.next.next != null) {
if (cur.next.val == cur.next.next.val) {
int x = cur.next.val;
while (cur.next != null && cur.next.val == x) {
cur.next = cur.next.next;
}
} else {
cur = cur.next;
}
}
return dummy.next;
}
public static void main(String[] args) {
ListNode n1 = new ListNode(1);
ListNode n2 = new ListNode(2);
n1.next = n2;
ListNode n3 = new ListNode(3);
n2.next = n3;
ListNode n4 = new ListNode(3);
n3.next = n4;
ListNode n5 = new ListNode(4);
n4.next = n5;
ListNode n6 = new ListNode(4);
n5.next = n6;
ListNode n7 = new ListNode(5);
n6.next = n7;
print(deleteDuplicates(n1));
n1 = new ListNode(1);
n2 = new ListNode(1);
n1.next = n2;
n3 = new ListNode(1);
n2.next = n3;
n4 = new ListNode(2);
n3.next = n4;
n5 = new ListNode(3);
n4.next = n5;
print(deleteDuplicates(n1));
n1 = new ListNode(1);
n2 = new ListNode(1);
n1.next = n2;
print(deleteDuplicates(n1));
}
public static class ListNode {
int val;
ListNode next;
ListNode() {
}
ListNode(int val) {
this.val = val;
}
ListNode(int val, ListNode next) {
this.val = val;
this.next = next;
}
}
private static void print(ListNode head) {
while (head != null) {
System.out.print(head.val + " ");
head = head.next;
}
System.out.println();
}
}
|
打印：
思路：
思路很简单。就是加一个哑节点。注意，不要思考过于复杂。