【LeetCode】Partition List

题目描述：

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

.
考察的链表处理。如果在原链表上改变指向来获取最终答案会非常麻烦。比较简洁的方式是将x左边和右边的节点分别连进新的链表，最后将两个链表拼起来。
代码：

class Solution {
public:
ListNode *partition(ListNode *head, int x) {
ListNode *a = new ListNode(0);
ListNode *b = new ListNode(0);
ListNode *small(a), *large(b);
while (head){
if (head->val < x){
small->next = head;
small = small->next;
}
else{
large->next = head;
large = large->next;
}
head = head->next;
}
large->next = NULL;
small->next = b->next;
return a->next;
}
};