LeetCode: Partition List 解题报告

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.

 /**
  * Definition for singly-linked list.
  * public class ListNode {
  *     int val;
  *     ListNode next;
  *     ListNode(int x) {
  *         val = x;
  *         next = null;
  *     }
  * }
  */
 public class Solution {
     public ListNode partition(ListNode head, int x) {
         if (head == null) {
             return null;
         }

         ListNode dummy = new ListNode(0);
         dummy.next = head;

         ListNode pre = dummy;
         ListNode cur = head;

         // Record the big list.
         ListNode bigDummy = new ListNode(0);
         ListNode bigTail = bigDummy;

         while (cur != null) {
             if (cur.val >= x) {
                 // Unlink the cur;
                 pre.next = cur.next;

                 // Add the cur to the tail of the new link.
                 bigTail.next = cur;
                 cur.next = null;

                 // Refresh the bigTail.
                 bigTail = cur;

                 // 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。
             } else {
                 pre = pre.next;
             }

             cur = pre.next;
         }

         // Link the Big linklist to the smaller one.
         pre.next = bigDummy.next;

         return dummy.next;
     }
 }

CODE:

https://github.com/yuzhangcmu/LeetCode_algorithm/blob/master/list/Partition.java