Partition List

Partition List

问题：

Given a linked list and a value x, partition it such that all nodes less than x come before nodes greater than or equal to x.

You should preserve the original relative order of the nodes in each of the two partitions.

思路：

　　画画图就OK了，基本的链表操作而已

我的代码：

public class Solution {
    public ListNode partition(ListNode head, int x) {
        if(head == null)    return head;
        ListNode dummy = new ListNode(-1);
        dummy.next = head;
        ListNode pre = dummy;
        while(head != null)
        {
            if(head.val >= x)   break;
            head = head.next;
            pre = pre.next;
        }
        if(head == null)    return dummy.next;
        ListNode tail = pre;
        while(head != null)
        {
            ListNode next = head.next;
            if(head.val < x)
            {
                tail.next = head.next;
                head.next = pre.next;
                pre.next = head;
                pre = head;
            }
            else
            {
                tail = tail.next;
            }
            head = next;            
        }
        return dummy.next;
    }
}

学习之处：

• 在链表中常用的插入一个节点的方法 带插入的节点node,node之前的oriPre 待插入位置的pre，三行情书，这三行仔细想想好像某种恋爱关系，不深入展开了。这种模式要记住，免得每一次还得画图，速度变慢了。

• origPre.next = cur.next;
  cur.next = pre.next;
  pre.next = cur;