96.Partition List-链表划分（容易题）

链表划分

1. 题目

   给定一个单链表和数值x，划分链表使得所有小于x的节点排在大于等于x的节点之前。

你应该保留两部分内链表节点原有的相对顺序。

1. 样例

   给定链表 1->4->3->2->5->2->null，并且 x=3

   返回 1->2->2->4->3->5->null

2. 题解

遍历找出小于x的节点组成新链表，后将两条链表合并。

/**
 * Definition for ListNode.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int val) {
 *         this.val = val;
 *         this.next = null;
 *     }
 * }
 */ 
public class Solution {
    /**
     * @param head: The first node of linked list.
     * @param x: an integer
     * @return: a ListNode 
     */
public ListNode partition(ListNode head, int x) {
if (head == null || head.next == null)
        {
            return head;
        }

        ListNode next = new ListNode(1);
next.next = head;
        ListNode smallHead = null;
        ListNode smallLast = null;
        ListNode bigHead = null;

while (next.next != null)
        {
if (next.next.val < x)
            {
if (smallHead == null)
                {
                    smallHead = next.next;
                    smallLast = smallHead;
                }
else
                {
                    smallLast.next = next.next;
                    smallLast = smallLast.next;
                }
next.next = next.next.next;
            }
else
            {
if (bigHead == null)
                {
                    bigHead = next.next;
                }
next = next.next;
            }
        }
if (smallHead != null)
        {
            smallLast.next = bigHead;
            return smallHead;
        }
        return bigHead;
    }
}

Last Update 2016.9.2