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83. 删除排序链表中的重复元素
题目描述
给定一个已排序的链表的头 head ， 删除所有重复的元素，使每个元素只出现一次 。返回 已排序的链表 。
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]
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**
难度： Easy
题解代码
javascript
/**
* 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) {
const arr = []
let p = head
while (p) {
if (p.val !== arr[arr.length - 1]) {
arr.push(p.val)
}
p = p.next
}
let p1 = head
for (let i = 0; i < arr.length; i++) {
p1.val = arr[i]
if (i === arr.length - 1) {
p1.next = null
} else {
p1 = p1.next
}
}
return head
};
var deleteDuplicates = function(head) {
const preHead = new ListNode(-1)
preHead.next = head
let p = preHead.next
while (p && p.next) {
while (p && p.next && p.val === p.next.val) {
p.next = p.next.next
}
p = p.next
}
return preHead.next
};
var deleteDuplicates = function (head) {
const map = new Map()
const prevHead = new ListNode()
prevHead.next = head
let p = prevHead
while (p && p.next) {
if (map.has(p.next.val)) {
p.next = p.next.next
} else {
map.set(p.next.val, true)
p = p.next
}
}
return prevHead.next
}