LeetCode——Remove Duplicates from Sorted List II

標籤：leetcode

Given a sorted linked list, delete all nodes that have duplicate numbers, leaving only distinct numbers from the original list.

For example,
Given 1->2->3->3->4->4->5, return 1->2->5.
Given 1->1->1->2->3, return 2->3.

原題連結：https://oj.leetcode.com/problems/remove-duplicates-from-sorted-list-ii/

題目：給定一個已排序的鏈表，重複資料刪除的元素，只留下源鏈表中唯一出現的那些元素。

思路：首先防止把前端節點刪除，構建一個虛擬前端節點。由於是已排序的，相等的元素就在相鄰的位置上，發現有相鄰的元素，則指標依次向後指，直到沒有相等的相鄰元素為止。

public ListNode deleteDuplicates(ListNode head) {if (head == null || head.next == null)return head;ListNode dummy = new ListNode(0);dummy.next = head;head = dummy;while (head.next != null && head.next.next != null) {if (head.next.val == head.next.next.val) {int val = head.next.val;while (head.next != null && head.next.val == val) {head.next = head.next.next;}} else {head = head.next;}}return dummy.next;}// Definition for singly-linked list.public class ListNode {int val;ListNode next;ListNode(int x) {val = x;next = null;}}