Open
Description
给你一个链表的头节点 head 和一个特定值 x ,请你对链表进行分隔,使得所有 小于 x 的节点都出现在 大于或等于 x 的节点之前。
你应当 保留 两个分区中每个节点的初始相对位置。
示例 1:
输入:head = [1,4,3,2,5,2], x = 3
输出:[1,2,2,4,3,5]
示例 2:
输入:head = [2,1], x = 2
输出:[1,2]
提示:
- 链表中节点的数目在范围 [0, 200] 内
- -100 <= Node.val <= 100
- -200 <= x <= 200
来源:力扣(LeetCode)
链接:https://leetcode-cn.com/problems/partition-list
著作权归领扣网络所有。商业转载请联系官方授权,非商业转载请注明出处。
const partition = (head, x) => {
let minHead = currMin = { val: -Infinity }
let prev = curr = { next: head }
while (curr.next) {
if (curr.next.val < x) {
currMin.next = curr.next
currMin = currMin.next
curr.next = curr.next.next
} else {
curr = curr.next
}
}
currMin.next = prev.next
return minHead.next
}