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

.

思路：分别用两个链表，一个保存比指定数小的元素，一个保存大于等于指定数的元素，最后把第二个链表接在第一个链表之后即可。

实现代码：

/**
* Definition for singly-linked list.
* struct ListNode {
*     int val;
*     ListNode *next;
*     ListNode(int x) : val(x), next(NULL) {}
* };
*/
class Solution {
public:
ListNode *partition(ListNode *head, int x) {
ListNode node1(0), node2(0);
ListNode *p1 = &node1, *p2 = &node2;
while (head) {
if (head->val < x)
p1 = p1->next = head;
else
p2 = p2->next = head;
head = head->next;
}
p2->next = NULL;
p1->next = node2.next;
return node1.next;
}
};