给定一个链表和一个特定值 x,对链表进行分隔,使得所有小于 x 的节点都在大于或等于 x 的节点之前。
你应当保留两个分区中每个节点的初始相对位置。
例如,
给定1->4->3->2->5->2 和 x = 3,
返回1->2->2->4->3->5.
详见:https://leetcode.com/problems/partition-list/description/
Java实现:
/**
* 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) {
if(head==null||head.next==null){
return head;
}
ListNode small = new ListNode(-1);
ListNode newSmall = small;
ListNode big = new ListNode(-1);
ListNode newBig = big;
while(head!=null){
if(head.val<x){
small.next = head;
small = small.next;
}else{
big.next = head;
big = big.next;
}
head = head.next;
}
big.next = null;
small.next = newBig.next;
return newSmall.next;
}
}
参考:https://www.cnblogs.com/springfor/p/3862392.html