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删除排序链表中的重复元素 II

雷亚荣 2022-02-12 阅读 59

问题:
给定一个已排序的链表的头 head , 删除原始链表中所有重复数字的节点,只留下不同的数字 。返回 已排序的链表 。

示例:
输入:head = [1,2,3,3,4,4,5]
输出:[1,2,5]

输入:head = [1,1,1,2,3]
输出:[2,3]

思想:
利用双端队列对链表节点进行添加和删除

代码:

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode() {}
 *     ListNode(int val) { this.val = val; }
 *     ListNode(int val, ListNode next) { this.val = val; this.next = next; }
 * }
 */
class Solution {
    LinkedList<Integer> repmax = new LinkedList<Integer>();
    public ListNode deleteDuplicates(ListNode head) {
        if(head == null) return null;
        ListNode cur = head;
        ListNode ans = head;
        int t = 101;
        while(cur!=null){
            if(!repmax.isEmpty()&&repmax.peekLast()!=cur.val&&cur.val!=t){
                
                repmax.addLast(cur.val);
                cur = cur.next;
                continue;
            }
            else if(!repmax.isEmpty()&&repmax.peekLast()==cur.val){
                t = cur.val;
                repmax.pollLast();
                cur = cur.next;
                continue;
            }
            else if(cur.val==t){
                t = cur.val;
                cur = cur.next;
                continue;
            } 
            repmax.addLast(cur.val);
            cur = cur.next;
        }
        if(repmax.isEmpty()) return null;
        while(!repmax.isEmpty()){
            head.val = repmax.peekFirst();
            repmax.pollFirst();
            if(repmax.isEmpty()) break;
            head = head.next;
        }
        head.next = null;
        return ans;

    }
}
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