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:找到第一个>=x的节点,记为p,p前一个节点记为s,向后遍历链表,如果节点的值小于x,就把该节点插入到s后面,同时s后移一位。
为了方便,使用了哨兵节点。
java:
/** * Definition for singly-linked list. * public class ListNode { * int val; * ListNode next; * ListNode(int x) { * val = x; * next = null; * } * } */ public class Solution { public ListNode partition(ListNode head, int x) { if(head==null||head.next==null) return head; ListNode h =new ListNode(-1); h.next=head; ListNode p=head; ListNode s=h; while(p!=null){ if(p.val>=x){ break; } s=s.next; p=p.next; } ListNode q=p; while(p!=null){ if(p.val<x){ ListNode tmp=p.next; q.next=tmp; p.next=s.next; s.next=p; s=s.next; } q=p; p=p.next; } return h.next; } }
c++:
/** * 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||!head->next) return head; ListNode * h =(ListNode *)malloc(sizeof(ListNode)); h->next=head; ListNode *p=head; ListNode *s=h; while(p){ if(p->val>=x){ break; } s=s->next; p=p->next; } ListNode *q=p; while(p){ if(p->val<x){ ListNode *tmp=p->next; q->next=tmp; p->next=s->next; s->next=p;
s=s->next;//后移一步 } q=p; p=p->next; } return h->next; } };
原文地址:http://blog.csdn.net/u010786672/article/details/41077197