LeetCode.分隔链表

给定一个仅包含0和1 、大小为rows x cols的二维二进制矩阵，找出只包含1的最大矩形，并返回其面积。给定一个链表和一个特定值x，对链表进行分隔，使得所有小于x的节点都在大
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:
  输入: head = 1->4->3->2->5->2, x = 3
  输出: 1->2->2->4->3->5

  解法：

  object PartitionLinkedList {
  
       def main(args: Array[String]): Unit = {
            println(partition(ListNode(1, ListNode(4, ListNode(3, ListNode(2, ListNode(5, ListNode(2, null)))))), 3))
       }
  
       def partition(head: ListNode, x: Int): ListNode = {
            if (head == null || head.next == null) return head
            var first: ListNode = ListNode(-1, null)
            val firstHead = first
            var second: ListNode = ListNode(-1, null)
            val secondHead: ListNode = second
  
            var cur = head
            while (cur != null) {
                 if (cur.x < x) {
                      first.next = cur
                      cur = cur.next
                      first = first.next
                      first.next = null
                 } else if (cur.x >= x) {
                      second.next = cur
                      cur = cur.next
                      second = second.next
                      second.next = null
                 }
            }
  
            first.next = secondHead.next
            firstHead.next
       }
  
       case class ListNode(x: Int, var next: ListNode)
  }