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96 lines
2.9 KiB
Markdown
96 lines
2.9 KiB
Markdown
# [2181. Merge Nodes in Between Zeros](https://leetcode.com/problems/merge-nodes-in-between-zeros/)
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## 题目
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You are given the `head` of a linked list, which contains a series of integers **separated** by `0`'s. The **beginning** and **end** of the linked list will have `Node.val == 0`.
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For **every** two consecutive `0`'s, **merge** all the nodes lying in between them into a single node whose value is the **sum** of all the merged nodes. The modified list should not contain any `0`'s.
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Return *the* `head` *of the modified linked list*.
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**Example 1:**
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```
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Input: head = [0,3,1,0,4,5,2,0]
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Output: [4,11]
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Explanation:
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The above figure represents the given linked list. The modified list contains
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- The sum of the nodes marked in green: 3 + 1 = 4.
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- The sum of the nodes marked in red: 4 + 5 + 2 = 11.
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```
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**Example 2:**
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```
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Input: head = [0,1,0,3,0,2,2,0]
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Output: [1,3,4]
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Explanation:
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The above figure represents the given linked list. The modified list contains
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- The sum of the nodes marked in green: 1 = 1.
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- The sum of the nodes marked in red: 3 = 3.
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- The sum of the nodes marked in yellow: 2 + 2 = 4.
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```
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**Constraints:**
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- The number of nodes in the list is in the range `[3, 2 * 10^5]`.
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- `0 <= Node.val <= 1000`
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- There are **no** two consecutive nodes with `Node.val == 0`.
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- The **beginning** and **end** of the linked list have `Node.val == 0`.
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## 题目大意
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给你一个链表的头节点 head ，该链表包含由 0 分隔开的一连串整数。链表的 开端 和 末尾 的节点都满足 Node.val == 0 。对于每两个相邻的 0 ，请你将它们之间的所有节点合并成一个节点，其值是所有已合并节点的值之和。然后将所有 0 移除，修改后的链表不应该含有任何 0 。
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返回修改后链表的头节点 head 。
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## 解题思路
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- 简单题。合并链表中两个值为 0 的节点。从头开始遍历链表，遇到节点值不为 0 的节点便累加；遇到节点值为 0 的节点，将累加值转换成结果链表要输出的节点值，然后继续遍历。
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## 代码
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```go
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package leetcode
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import (
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"github.com/halfrost/LeetCode-Go/structures"
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)
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// ListNode define
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type ListNode = structures.ListNode
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/**
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* Definition for singly-linked list.
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* type ListNode struct {
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* Val int
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* Next *ListNode
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* }
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*/
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func mergeNodes(head *ListNode) *ListNode {
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res := &ListNode{}
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h := res
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if head.Next == nil {
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return &structures.ListNode{}
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}
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cur := head
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sum := 0
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for cur.Next != nil {
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if cur.Next.Val != 0 {
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sum += cur.Next.Val
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} else {
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h.Next = &ListNode{Val: sum, Next: nil}
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h = h.Next
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sum = 0
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}
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cur = cur.Next
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}
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return res.Next
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}
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``` |