Leetcode: 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.

分析：可以将小于x的node放在一个链表中，大于等于x的node放在另一个链表中，然后将两个链表首尾相连。

class Solution {
public:
    ListNode *partition(ListNode *head, int x) {
        if(head == NULL) return head;
        
        ListNode *shead = new ListNode(-1);
        ListNode *bhead = new ListNode(-1);
        
        ListNode *ps = shead, *pb = bhead;
        while(head){
            ListNode *tmp = head;
            head = head->next;
            tmp->next = NULL;
            if(tmp->val < x){
                ps->next = tmp;
                ps = ps->next;
            }else{
                pb->next = tmp;
                pb = pb->next;
            }
        }
        ps->next = bhead->next;
        
        return shead->next;
    }
};