Partition List

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.

For example,

Given 1->4->3->2->5->2 and x = 3,

return 1->2->2->4->3->5.

SOLUTION 1:

注意使用Dummynode来记录各个链条的头节点的前一个节点。这样我们可以轻松找回头节点。

1. Go Through the link, find the nodes which are bigger than N, create a new link.

2. After 1 done, just link the two links.

1 /** 2  * Definition for singly-linked list. 3  * public class ListNode { 4  *     int val; 5  *     ListNode next; 6  *     ListNode(int x) { 7  *         val = x; 8  *         next = null; 9  *     }10  * }11  */12 public class Solution {13     public ListNode partition(ListNode head, int x) {14         if (head == null) {15             return null;16         }17         18         ListNode dummy = new ListNode(0);19         dummy.next = head;20         21         ListNode pre = dummy;22         ListNode cur = head;23         24         // Record the big list.25         ListNode bigDummy = new ListNode(0);26         ListNode bigTail = bigDummy;27         28         while (cur != null) {29             if (cur.val >= x) {30                 // Unlink the cur;31                 pre.next = cur.next;32                33                 // Add the cur to the tail of the new link.34                 bigTail.next = cur;35                 cur.next = null;36                37                 // Refresh the bigTail.38                 bigTail = cur;39                40                 // 移除了一个元素的时候，pre不需要修改，因为cur已经移动到下一个位置了。41             } else {42                 pre = pre.next;43             }44             45             cur = pre.next;46         }47         48         // Link the Big linklist to the smaller one.49         pre.next = bigDummy.next;50         51         return dummy.next;52     }53 }

CODE: