LeetCode OJ:Partition List

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.

算法思想：

两个指针，s将小于x的串起来，e将大于等于x的串起来，最后将s的next指向e，注意处理head指针就好了

/** * 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)return NULL;        ListNode *p,*q,*e,*s,*start,*end;        start=s=new ListNode(0);        end=e=new ListNode(0);        p=head;        while(p){            if(p->val<x){                s->next=p;                s=p;            }            else{                e->next=p;                e=p;            }            p=p->next;        }        e->next=NULL;        s->next=end->next;        return start->next;    }};