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面试题 02.04. 分割链表

English Version

题目描述

编写程序以 x 为基准分割链表，使得所有小于 x 的节点排在大于或等于 x 的节点之前。如果链表中包含 x，x 只需出现在小于 x 的元素之后(如下所示)。分割元素 x 只需处于“右半部分”即可，其不需要被置于左右两部分之间。

示例:

输入: head = 3->5->8->5->10->2->1, x = 5
输出: 3->1->2->10->5->5->8

解法

创建两个链表，一个存放小于 x 的节点，另一个存放大于等于 x 的节点，之后进行拼接即可。

Python3

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, x):
#         self.val = x
#         self.next = None

class Solution:
    def partition(self, head: ListNode, x: int) -> ListNode:
        if head is None or head.next is None:
            return head
        left, right = ListNode(-1), ListNode(-1)
        p, q = left, right
        while head:
            t = head.next
            head.next = None
            if head.val < x:
                p.next = head
                p = p.next
            else:
                q.next = head
                q = q.next
            head = t
        p.next = right.next
        return left.next

Java

/**
 * 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 left = new ListNode(-1);
        ListNode right = new ListNode(-1);
        ListNode p = left, q = right;
        while (head != null) {
            ListNode t = head.next;
            head.next = null;
            if (head.val < x) {
                p.next = head;
                p = p.next;
            } else {
                q.next = head;
                q = q.next;
            }
            head = t;
        }
        p.next = right.next;
        return left.next;
    }
}

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