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.
给定一个链表和一个整数n,要求把小于n的放到链表的前面,大于等于n的放到链表的后面,不要打乱原有的顺序。我们可以借助两个辅助节点,smallNode和bigNode分别将链表分为值小于n得一段和值大于等于n的一段,最后将两段连接起来。代码如下:
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.
给定一个链表和一个整数n,要求把小于n的放到链表的前面,大于等于n的放到链表的后面,不要打乱原有的顺序。我们可以借助两个辅助节点,smallNode和bigNode分别将链表分为值小于n得一段和值大于等于n的一段,最后将两段连接起来。代码如下:
/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode(int x) { val = x; }
* }
*/
public class Solution {
public ListNode partition(ListNode head, int x) {
if(head == null) return head;
ListNode smallNode = new ListNode(0);
ListNode helperS = smallNode;
ListNode bigNode = new ListNode(0);
ListNode helperB = bigNode;
while(head != null) {
if(head.val < x) {
smallNode.next = head;
smallNode = smallNode.next;
} else {
bigNode.next = head;
bigNode = bigNode.next;
}
head = head.next;
}
bigNode.next = null;
smallNode.next = helperB.next;
return helperS.next;
}
}
本文介绍了一种链表分区算法,该算法将链表中小于给定值n的节点置于其前面,而大于等于n的节点置于其后,同时保持原有相对顺序不变。通过使用两个辅助节点smallNode和bigNode来实现这一目标。
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