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.
/**
* 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) {
ListNode lin = head;
ListNode small = new ListNode(0);
ListNode hs = small;
ListNode large = new ListNode(0);
ListNode hl = large;
while(lin!=null){
ListNode ln = new ListNode(lin.val);
if(lin.val<x){
small.next = ln;
small = small.next;
}else{
large.next = ln;
large = large.next;
}
lin = lin.next;
}
small.next = hl.next;
return hs.next;
}
}
原文地址:http://blog.csdn.net/guorudi/article/details/41120505