83. 删除排序链表中的重复元素

题目来源：83. 删除排序链表中的重复元素

存在一个按升序排列的链表，给你这个链表的头节点 head ，请你删除所有重复的元素，使每个元素 只出现一次 。

返回同样按升序排列的结果链表。

方法一：一次遍历

/**
 * Definition for singly-linked list.
 * function ListNode(val, next) {
 *     this.val = (val===undefined ? 0 : val)
 *     this.next = (next===undefined ? null : next)
 * }
 */
/**
 * @param {ListNode} head
 * @return {ListNode}
 */
var deleteDuplicates = function(head) {
  if(!head){
    return head;
  }
  let cur = head;
  while(cur.next){
    if(cur.val === cur.next.val){
      cur.next = cur.next.next;
    }else{
      cur = cur.next;
    }
  }
  return head;
};

方法二：哈希

/**
 * Definition for singly-linked list.
 * function ListNode(val, next) {
 *     this.val = (val===undefined ? 0 : val)
 *     this.next = (next===undefined ? null : next)
 * }
 */
/**
 * @param {ListNode} head
 * @return {ListNode}
 */
var deleteDuplicates = function(head) {
  let cur = head;
  let pre = cur;
  let hash = new Map();
  while(cur){
    let val = cur.val;
    if(!hash.has(val)){
      hash.set(val, 1);
      pre = cur;
    }else{
      pre.next = cur.next;
    }
    cur = cur.next;
  }
  return head;
};

示例 1：

输入：head = [1,1,2]
输出：[1,2]

示例 2：

输入：head = [1,1,2,3,3]
输出：[1,2,3]

 python3

# Definition for singly-linked list.
class ListNode:
    def __init__(self, val=0, next=None):
        self.val = val
        self.next = next
class Solution:
    def deleteDuplicates(self, head: ListNode) -> ListNode:
      if not head:
        return head
      
      cur = head
      while cur:
        if cur.val == cur.next.val:
          cur.next = cur.next.next
        else:
          cur = cur.next

      return head

 

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