Description
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.
Solution
Two-pointers
比較簡單的做法是把節(jié)點移動到兩個新鏈表中拱镐，最后把兩個新鏈表拼起來艘款。注意不要形成環(huán)。
/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode(int x) { val = x; }
* }
*/
class Solution {
public ListNode partition(ListNode head, int x) {
ListNode smallHead = new ListNode(0);
ListNode largeHead = new ListNode(0);
ListNode smallTail = smallHead;
ListNode largeTail = largeHead;
while (head != null) {
if (head.val < x) {
smallTail.next = head;
smallTail = smallTail.next;
} else {
largeTail.next = head;
largeTail = largeTail.next;
}
head = head.next;
}
smallTail.next = largeHead.next;
largeTail.next = null; // in case there's a loop
return smallHead.next;
}
}
