[LeetCode 题解]: 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.

题意：

给定一个链表以及一个数x。

要求：

将队列中所有小于x的节点放到大于或等于x的节点之前。

思路：

根据题意，很容易想到利用两个链表：

lower按照先后顺序存放小于x的节点；

higher存放大于等于x的节点。

最后合并分割后的链表，按照lower在前，higher在后的顺序。

class Solution {

public:

    ListNode *partition(ListNode *head, int x) {

        if(head==NULL || head->next==NULL) return head;

        // divide list into two parts:

        // lower part with all nodes whose value is less than x

        // higher part with all nodes whose value is more than x

        ListNode *lower = new ListNode(-1); 

        ListNode *higher = new ListNode(-1);

        ListNode *ltail = lower;

        ListNode *htail = higher;

        ListNode *tmp = head;

        while(tmp){

            if(tmp->val < x){

                ltail->next = tmp;

                ltail = ltail->next;

            }else{

                htail->next = tmp;

                htail = htail->next;

            }

            tmp = tmp->next; 

        }

        htail->next=NULL; // set end of a list

        ltail->next=higher->next; // append higher part to the tail of lower part

        return lower->next;        

    }

};