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。
示例 1:
输入: 1->2->3->3->4->4->5
输出: 1->2->5
示例 2:
输入: 1->1->1->2->3
输出: 2->3
来源:力扣(LeetCode)
链接:https://leetcode-cn.com/problems/remove-duplicates-from-sorted-list-ii
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1 public class _82 { 2 public ListNode deleteDuplicates(ListNode head) { 3 if (head == null || head.next == null) return head; // 有0个或1个节点 4 if (head.next.next == null && head.val == head.next.val) return null; // 有2个值相同节点 5 ListNode p = head, q = p.next; 6 List<ListNode> noRepeat = new ArrayList<>(); 7 if (p.val != q.val) { 8 noRepeat.add(p); 9 } 10 // 扫一遍将不重复的数字记录下来 11 while (q != null){ 12 if (p.val != q.val && q.next != null && q.val != q.next.val) 13 noRepeat.add(q); 14 if (q.next == null && p.val != q.val){ 15 noRepeat.add(q); 16 } 17 p = p.next; 18 q = p.next; 19 } 20 for (int i = 0; i < noRepeat.size()-1; i++){ 21 noRepeat.get(i).next = noRepeat.get(i+1); 22 } 23 if (noRepeat.size() <= 0) 24 return null; 25 noRepeat.get(noRepeat.size()-1).next = null; 26 return noRepeat.get(0); 27 } 28 29 public static ListNode create(int[] elems){ 30 if (elems == null) return null; 31 ListNode head = new ListNode(-1); 32 ListNode p = head; 33 for (int e : elems){ 34 ListNode listNode = new ListNode(e); 35 listNode.next = null; 36 p.next = listNode; 37 p = p.next; 38 } 39 return head.next; 40 } 41 42 public static void main(String[] args) { 43 int[] elems = {1,1,1,2,2,3,3,4,5,6,7}; 44 ListNode head = create(elems); 45 ListNode listNode = new _82().deleteDuplicates(head); 46 System.out.println(); 47 while (listNode != null) { 48 System.out.print(listNode.val+", "); 49 listNode = listNode.next; 50 } 51 } 52 }
原文地址:https://www.cnblogs.com/yfs123456/p/11545714.html
时间: 2024-10-06 02:24:59