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86. 分隔链表

English Version

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

方法一：模拟

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。

时间复杂度 $O(n)，空间复杂度 $O(1)$。其中 $n$ 是原链表的长度。

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: Optional[ListNode], x: int) -> Optional[ListNode]:
        d1, d2 = ListNode(), ListNode()
        t1, t2 = d1, d2
        while head:
            if head.val < x:
                t1.next = head
                t1 = t1.next
            else:
                t2.next = head
                t2 = t2.next
            head = head.next
        t1.next = d2.next
        t2.next = None
        return d1.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 d1 = new ListNode();
        ListNode d2 = new ListNode();
        ListNode t1 = d1, t2 = d2;
        while (head != null) {
            if (head.val < x) {
                t1.next = head;
                t1 = t1.next;
            } else {
                t2.next = head;
                t2 = t2.next;
            }
            head = head.next;
        }
        t1.next = d2.next;
        t2.next = null;
        return d1.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* d1 = new ListNode();
        ListNode* d2 = new ListNode();
        ListNode* t1 = d1;
        ListNode* t2 = d2;
        while (head) {
            if (head->val < x) {
                t1->next = head;
                t1 = t1->next;
            } else {
                t2->next = head;
                t2 = t2->next;
            }
            head = head->next;
        }
        t1->next = d2->next;
        t2->next = nullptr;
        return d1->next;
    }
};

Go

/**
 * Definition for singly-linked list.
 * type ListNode struct {
 *     Val int
 *     Next *ListNode
 * }
 */
func partition(head *ListNode, x int) *ListNode {
	d1, d2 := &ListNode{}, &ListNode{}
	t1, t2 := d1, d2
	for head != nil {
		if head.Val < x {
			t1.Next = head
			t1 = t1.Next
		} else {
			t2.Next = head
			t2 = t2.Next
		}
		head = head.Next
	}
	t1.Next = d2.Next
	t2.Next = nil
	return d1.Next
}

JavaScript

/**
 * Definition for singly-linked list.
 * function ListNode(val, next) {
 *     this.val = (val===undefined ? 0 : val)
 *     this.next = (next===undefined ? null : next)
 * }
 */
/**
 * @param {ListNode} head
 * @param {number} x
 * @return {ListNode}
 */
var partition = function (head, x) {
    const d1 = new ListNode();
    const d2 = new ListNode();
    let t1 = d1,
        t2 = d2;
    while (head) {
        if (head.val < x) {
            t1.next = head;
            t1 = t1.next;
        } else {
            t2.next = head;
            t2 = t2.next;
        }
        head = head.next;
    }
    t1.next = d2.next;
    t2.next = null;
    return d1.next;
};

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