LeetCode OJ：Partition List（分割链表）

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

.

将链表分割成两部分，大于某个数字的在左侧，小于等于某个数字的在右侧，用的方法比较蠢可能就是遍历一次分成两个链表，然后再将它们接起来。具体代码如下所示：

class Solution {
public:
ListNode* partition(ListNode* head, int x) {
ListNode * helper1 = new ListNode(INT_MIN);
ListNode * helper2 = new ListNode(INT_MIN);
ListNode * p1 = helper1;
ListNode * p2 = helper2;
while(head){
if(head->val < x){
p1->next = head;
head = head->next;
p1 = p1->next;
p1->next = NULL;
}else{
p2->next = head;
head = head->next;
p2 = p2->next;
p2->next = NULL;
}
}
p1->next = helper2->next;
return helper1->next;
}
};