README.md

1669. 合并两个链表

English Version

题目描述

给你两个链表 list1 和 list2 ，它们包含的元素分别为 n 个和 m 个。

请你将 list1 中下标从 a 到 b 的全部节点都删除，并将list2 接在被删除节点的位置。

下图中蓝色边和节点展示了操作后的结果：

请你返回结果链表的头指针。

 

示例 1：

输入：list1 = [0,1,2,3,4,5], a = 3, b = 4, list2 = [1000000,1000001,1000002]
输出：[0,1,2,1000000,1000001,1000002,5]
解释：我们删除 list1 中下标为 3 和 4 的两个节点，并将 list2 接在该位置。上图中蓝色的边和节点为答案链表。

示例 2：

输入：list1 = [0,1,2,3,4,5,6], a = 2, b = 5, list2 = [1000000,1000001,1000002,1000003,1000004]
输出：[0,1,1000000,1000001,1000002,1000003,1000004,6]
解释：上图中蓝色的边和节点为答案链表。

 

提示：

• 3 <= list1.length <= 104
• 1 <= a <= b < list1.length - 1
• 1 <= list2.length <= 104

解法

方法一：模拟

直接模拟题目中的操作即可。

时间复杂度 $O(n)$，空间复杂度 $O(1)$。其中 $n$ 为链表 list1 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 mergeInBetween(
        self, list1: ListNode, a: int, b: int, list2: ListNode
    ) -> ListNode:
        p = q = list1
        for _ in range(a - 1):
            p = p.next
        for _ in range(b + 1):
            q = q.next
        t = list2
        while t.next:
            t = t.next
        t.next = q
        p.next = list2
        return list1

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 mergeInBetween(ListNode list1, int a, int b, ListNode list2) {
        ListNode p = list1, q = list1;
        for (int i = 0; i < a - 1; ++i) {
            p = p.next;
        }
        for (int i = 0; i < b + 1; ++i) {
            q = q.next;
        }
        ListNode t = list2;
        while (t.next != null) {
            t = t.next;
        }
        t.next = q;
        p.next = list2;
        return list1;
    }
}

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* mergeInBetween(ListNode* list1, int a, int b, ListNode* list2) {
        auto p = list1, q = list1;
        for (int i = 0; i < a - 1; ++i) {
            p = p->next;
        }
        for (int i = 0; i < b + 1; ++i) {
            q = q->next;
        }
        auto t = list2;
        while (t->next) {
            t = t->next;
        }
        t->next = q;
        p->next = list2;
        return list1;
    }
};

Go

/**
 * Definition for singly-linked list.
 * type ListNode struct {
 *     Val int
 *     Next *ListNode
 * }
 */
func mergeInBetween(list1 *ListNode, a int, b int, list2 *ListNode) *ListNode {
	p, q := list1, list1
	for i := 0; i < a-1; i++ {
		p = p.Next
	}
	for i := 0; i < b+1; i++ {
		q = q.Next
	}
	t := list2
	for t.Next != nil {
		t = t.Next
	}
	t.Next = q
	p.Next = list2
	return list1
}

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