[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(-);
        ListNode *higher = new ListNode(-);
        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;
    }
};