题解 | #删除链表中重复的结点#

/* public class ListNode { int val; ListNode next = null;

ListNode(int val) {
    this.val = val;
}

} */

import java.util.*;

public class Solution {

public ListNode deleteDuplication(ListNode pHead) {
//先使用散列表HashMap记录无重复数字的节点
//以{1，2，2，3，3，4}为例，得到{1：true, 2:false, 3:false, 4:true}
    HashMap<Integer, Boolean> map = new HashMap<>();
    ListNode cur = pHead;
    while(cur != null){
        if(map.containsKey(cur.val)){
            map.put(cur.val, false);
        }else{
            map.put(cur.val, true);
        }
        cur = cur.next;
    }
    //设置遍历指针cur初始位于pHead, 以及记录指针tmp
    //设置头节点head，记录初始tmp指针的位置，（同时是为了防止初始pHead.val为重复数值出错）
    //head.next = pHead;
    cur = pHead;
    ListNode tmp = new ListNode(-1);
    tmp.next = pHead;
    ListNode head = tmp;
    while(cur != null){
        
        if(map.get(cur.val).equals(true)){
            tmp.next = cur;
            tmp = cur;
            cur = cur.next;
            
        }else{
            cur = cur.next;
        }
    }
    //一下是针对{1，1}出错所做的修改
    tmp.next = cur;
    tmp = cur;

    return head.next;
}

}