leetCode 86.Partition List(分区链表) 解题思路和方法

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.

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
public class Solution {
    public ListNode partition(ListNode head, int x) {
    	/**
    	 * 思路是将list按X分成两段
    	 * 小于的连接p
    	 * 大于的连接q
    	 * 最后合并p和q即可
    	 */
        ListNode p = new ListNode(0);
        ListNode pHead = p;
        ListNode q = new ListNode(0);
        ListNode qHead = q;
        //遍历
        while(head != null){
            if(head.val < x){//<x成一组
                p.next = head;
                p = p.next;
            }else{//>=x成一组
                q.next = head;
                q = q.next;
            }
            head = head.next;
        }
        p.next = qHead.next;
        q.next = null;//斩断后面的连接
        return pHead.next;
    }
}

leetCode 86.Partition List(分区链表) 解题思路和方法

原文：http://blog.csdn.net/xygy8860/article/details/47053445