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
.
思考:遍历链表,小于x的存于链表head1,大于等于的存于链表head2,合并head1,head2.
/** * 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 *p=head; ListNode *head1=NULL; ListNode *head2=NULL; ListNode *p1,*p2; while(p) { ListNode *node=new ListNode(p->val); if(node->val<x) { if(head1==NULL) { head1=p1=node; } else { p1->next=node; p1=node; } } else { if(head2==NULL) { head2=p2=node; } else { p2->next=node; p2=node; } } p=p->next; } p1->next=head2; if(head1==NULL) return head2; return head1; } };