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题目：
给定一个单链表和数值x，划分链表使得所有小于x的节点排在大于等于x的节点之前。
你应该保留两部分内链表节点原有的相对顺序。
给定链表 1->4->3->2->5->2->null，并且 x=3
返回 1->2->2->4->3->5->null
解题：
上面返回的结果好多不对的应该是： 1->2->2->3->4->5->null
想了好久不知道怎么搞，九章看到的程序，竟然搞两个链表链接起来。。。
Java程序：
/** * Definition for ListNode. * public class ListNode { * int val; * ListNode next; * ListNode(int val) { * this.val = val; * this.next = null; * } * } */ public class Solution { /** * @param head: The first node of linked list. * @param x: an integer * @return: a ListNode */ public ListNode partition(ListNode head, int x) { // write your code here if(head == null) return null; ListNode leftDummy = new ListNode(0); ListNode rightDummy = new ListNode(0); ListNode left = leftDummy, right = rightDummy; while (head != null) { if (head.val < x) { left.next = head; left = head; } else { right.next = head; right = head; } head = head.next; } right.next = null; left.next = rightDummy.next; return leftDummy.next; } }
总耗时: 1573 ms
Python程序：
""" Definition of ListNode class ListNode(object): def __init__(self, val, next=None): self.val = val self.next = next """ class Solution: """ @param head: The first node of linked list. @param x: an integer @return: a ListNode """ def partition(self, head, x): # write your code here if head == None: return head lefthead = ListNode(0) righthead = ListNode(0) left = lefthead right = righthead while head!=None: if head.val<x: left.next = head left = left.next else: right.next = head right = right.next head = head.next right.next = None left.next = righthead.next return lefthead.next
总耗时: 517 ms
和Java的有点小区别但是没有影响的。
lintcode 容易题：Partition List 链表划分
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原文地址：http://www.cnblogs.com/theskulls/p/4889916.html
