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面试题 02.01. 移除重复节点

English Version

题目描述

编写代码，移除未排序链表中的重复节点。保留最开始出现的节点。

示例1:

 输入：[1, 2, 3, 3, 2, 1]
 输出：[1, 2, 3]

示例2:

 输入：[1, 1, 1, 1, 2]
 输出：[1, 2]

提示：

1. 链表长度在[0, 20000]范围内。
2. 链表元素在[0, 20000]范围内。

进阶：

如果不得使用临时缓冲区，该怎么解决？

解法

Python3

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, x):
#         self.val = x
#         self.next = None

class Solution:
    def removeDuplicateNodes(self, head: ListNode) -> ListNode:
        if head is None or head.next is None:
            return head
        cache = set()
        cache.add(head.val)
        cur, p = head, head.next
        while p:
            if p.val not in cache:
                cur.next = p
                cur = cur.next
                cache.add(p.val)
            p = p.next
        cur.next = None
        return head

Java

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
class Solution {
    public ListNode removeDuplicateNodes(ListNode head) {
        if (head == null || head.next == null) {
            return head;
        }
        Set<Integer> s = new HashSet<>();
        s.add(head.val);
        ListNode p = head.next, cur = head;
        while (p != null) {
            if (!s.contains(p.val)) {
                cur.next = p;
                cur = cur.next;
                s.add(p.val);
            }
            p = p.next;
        }
        cur.next = null;
        return head;

    }
}

JavaScript

/**
 * Definition for singly-linked list.
 * function ListNode(val) {
 *     this.val = val;
 *     this.next = null;
 * }
 */
/**
 * @param {ListNode} head
 * @return {ListNode}
 */
 var removeDuplicateNodes = function(head) {
    if (head == null || head.next == null) return head;
    const cache = new Set([]);
    cache.add(head.val);
    let cur = head, fast = head.next;
    while (fast !== null) {
        if (!cache.has(fast.val)) {
            cur.next = fast;
            cur = cur.next;
            cache.add(fast.val);
        }
        fast = fast.next;
    }
    cur.next = null;
    return head;
};

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