[leetcode] 86. Partition List 解题报告

题目链接：https://leetcode.com/problems/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：悲了催的昨天考驾照permit差点就过了，３０题可以错６道，我错了七道。最后两题被双绝杀．

这题可以用将链表分割成两个链表，一个保存小于x的数，一个保存大于等于x的数．最后将两个链表合并起来．一般这种链表的题目如果链表的头结点不确定的情况下使用虚拟头结点会方便很多，但是最后一定要将虚拟头结点的内存释放掉，内存泄露是非常严重的问题．

代码如下：

/**
* 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) {
if(!head || !head->next) return head;
ListNode *L1=new ListNode(0), *L2=new ListNode(0), *p = head;
ListNode *m = L1, *n = L2;
while(p)
{
if(p->val < x) m = m->next = p;
else n = n->next = p;
p = p->next;
}
n->next = NULL;
m->next = L2->next;
head = L1->next;
delete L1, L2;
return head;
}
};