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;
}
};