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;    }};