LeetCode: Partition List [086]

【题目】

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

.

【题意】

给定一个链表和一个数值x，将链表中的值按x进行划分，小于x的在前，大于等于x的在后。两部分中节点的之间的相对位置与在原链表中时相同

【思路】

先将链表按x分裂成两个链表，一个链表中的值小于x, 另一个链表中的值大于等于x

然后在将两个链表链接起来。

【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) {
if(head==NULL)return head;

ListNode*headLess=NULL;
ListNode*tailLess=NULL;
ListNode*headGreater=NULL;
ListNode*tailGreater=NULL;
ListNode*pointer=head;
while(pointer){
if(pointer->val<x){
if(tailLess==NULL)headLess=pointer;
else tailLess->next=pointer;
tailLess=pointer;
pointer=pointer->next;
tailLess->next=NULL;
}
else{
if(tailGreater==NULL)headGreater=pointer;
else tailGreater->next=pointer;
tailGreater=pointer;
pointer=pointer->next;
tailGreater->next=NULL;
}
}
//合并两个链表
if(tailLess){
head=headLess;
tailLess->next=headGreater;
}
else head=headGreater;

return head;
}
};