leetcode-链表-分割链表

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。

输入： head = [1,4,3,2,5,2], x = 3
输出：[1,2,2,4,3,5]

第一种解法

/**
 * Definition for singly-linked list.
 * type ListNode struct {
 *     Val int
 *     Next *ListNode
 * }
 */
func partition(head *ListNode, x int) *ListNode {
    small := &ListNode{}
    smallHead := small

    large := &ListNode{}
    largeHead := large
    for head != nil {
        if head.Val < x {
            small.Next = head
            small = small.Next
        } else {
            large.Next = head
            large = large.Next 
        }
        head = head.Next
    }
    large.Next = nil
    small.Next = largeHead.Next
    return smallHead.Next
}