86. 分隔链表 (opens new window)

English Version (opens new window)

题目描述

给你一个链表的头节点 head 和一个特定值x ，请你对链表进行分隔，使得所有 小于 x 的节点都出现在 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。

示例 1：

输入：head = [1,4,3,2,5,2], x = 3
输出：[1,2,2,4,3,5]

示例 2：

输入：head = [2,1], x = 2
输出：[1,2]

提示：

• 链表中节点的数目在范围 [0, 200] 内
• -100 <= Node.val <= 100
• -200 <= x <= 200

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, val=0, next=None):
#         self.val = val
#         self.next = next
class Solution:
    def partition(self, head: ListNode, x: int) -> ListNode:
        l1, l2 = ListNode(), ListNode()
        cur1, cur2 = l1, l2
        while head:
            if head.val < x:
                cur1.next = head
                cur1 = cur1.next
            else:
                cur2.next = head
                cur2 = cur2.next
            head = head.next
        cur1.next = l2.next
        cur2.next = None
        return l1.next

Java

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode() {}
 *     ListNode(int val) { this.val = val; }
 *     ListNode(int val, ListNode next) { this.val = val; this.next = next; }
 * }
 */
class Solution {
    public ListNode partition(ListNode head, int x) {
        ListNode l1 = new ListNode(0);
        ListNode l2 = new ListNode(0);
        ListNode cur1 = l1, cur2 = l2;
        while (head != null) {
            if (head.val < x) {
                cur1.next = head;
                cur1 = cur1.next;
            } else {
                cur2.next = head;
                cur2 = cur2.next;
            }
            head = head.next;
        }
        cur1.next = l2.next;
        cur2.next = null;
        return l1.next;
    }
}

C++

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode() : val(0), next(nullptr) {}
 *     ListNode(int x) : val(x), next(nullptr) {}
 *     ListNode(int x, ListNode *next) : val(x), next(next) {}
 * };
 */
class Solution {
public:
    ListNode* partition(ListNode* head, int x) {
        ListNode* l1 = new ListNode();
        ListNode* l2 = new ListNode();
        ListNode* cur1 = l1;
        ListNode* cur2 = l2;
        while (head != nullptr) {
            if (head->val < x) {
                cur1->next = head;
                cur1 = cur1->next;
            } else {
                cur2->next = head;
                cur2 = cur2->next;
            }
            head = head->next;
        }
        cur1->next = l2->next;
        cur2->next = nullptr;
        return l1->next;
    }
};

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