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.
最简答的，扫描两次链表
第一次找到所有比x小的节点，并保存下来
第二次找到所有比x大的节点，并保存下来
然后把他们两连起来
1 /** 2 * Definition for singly-linked list. 3 * struct ListNode { 4 * int val; 5 * ListNode *next; 6 * ListNode(int x) : val(x), next(NULL) {} 7 * }; 8 */ 9 class Solution { 10 public: 11 ListNode *partition(ListNode *head, int x) { 12 13 ListNode *p=head; 14 ListNode *p0=NULL; 15 ListNode *head1=NULL; 16 ListNode *head2=NULL; 17 while(p!=NULL) 18 { 19 ListNode *tmp; 20 if(p->val<x) 21 { 22 tmp=new ListNode(p->val); 23 if(p0==NULL) head1=tmp; 24 else p0->next=tmp; 25 p0=tmp; 26 } 27 28 p=p->next; 29 } 30 31 ListNode *head1End=p0; 32 p=head; 33 p0=NULL; 34 while(p!=NULL) 35 { 36 ListNode *tmp; 37 if(p->val>=x) 38 { 39 tmp=new ListNode(p->val); 40 if(p0==NULL) head2=tmp; 41 else p0->next=tmp; 42 p0=tmp; 43 } 44 p=p->next; 45 } 46 47 48 if(head1End!=NULL) head1End->next=head2; 49 else head1=head2; 50 51 return head1; 52 } 53 };
第二种思路，与第一种方法类似，只是不新建链表进行存储
1 /** 2 * Definition for singly-linked list. 3 * struct ListNode { 4 * int val; 5 * ListNode *next; 6 * ListNode(int x) : val(x), next(NULL) {} 7 * }; 8 */ 9 class Solution { 10 public: 11 ListNode *partition(ListNode *head, int x) { 12 13 if(head==NULL) return NULL; 14 15 ListNode *p=head; 16 17 ListNode *head1=NULL; 18 ListNode *head2=NULL; 19 20 ListNode *p0=NULL; 21 ListNode *p1=NULL; 22 23 24 25 while(p!=NULL) 26 { 27 if(p->val<x) 28 { 29 if(p0==NULL) head1=p; 30 else p0->next=p; 31 p0=p; 32 } 33 else 34 { 35 if(p1==NULL) head2=p; 36 else p1->next=p; 37 p1=p; 38 } 39 40 p=p->next; 41 } 42 43 if(p1!=NULL) p1->next=NULL; 44 if(p0!=NULL) p0->next=head2; 45 else head1=head2; 46 47 return head1; 48 } 49 };