2181. 合并零之间的节点

题目描述

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 。

返回修改后链表的头节点 head 。

示例 1：

输入：head = [0,3,1,0,4,5,2,0]
输出：[4,11]
解释：
上图表示输入的链表。修改后的链表包含：
- 标记为绿色的节点之和：3 + 1 = 4
- 标记为红色的节点之和：4 + 5 + 2 = 11

示例 2：

输入：head = [0,1,0,3,0,2,2,0]
输出：[1,3,4]
解释：
上图表示输入的链表。修改后的链表包含：
- 标记为绿色的节点之和：1 = 1
- 标记为红色的节点之和：3 = 3
- 标记为黄色的节点之和：2 + 2 = 4

提示：

列表中的节点数目在范围 [3, 2 * 10⁵] 内
0 <= Node.val <= 1000
不存在连续两个 Node.val == 0 的节点
链表的开端和末尾节点都满足 Node.val == 0

解法

方法一

Python3JavaC++GoTypeScriptRustC

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
class Solution:
    def mergeNodes(self, head: Optional[ListNode]) -> Optional[ListNode]:
        dummy = tail = ListNode()
        s = 0
        cur = head.next
        while cur:
            if cur.val != 0:
                s += cur.val
            else:
                tail.next = ListNode(s)
                tail = tail.next
                s = 0
            cur = cur.next
        return dummy.next

/**
 * 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 mergeNodes(ListNode head) {
        ListNode dummy = new ListNode();
        int s = 0;
        ListNode tail = dummy;
        for (ListNode cur = head.next; cur != null; cur = cur.next) {
            if (cur.val != 0) {
                s += cur.val;
            } else {
                tail.next = new ListNode(s);
                tail = tail.next;
                s = 0;
            }
        }
        return dummy.next;
    }
}

/**
 * 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* mergeNodes(ListNode* head) {
        ListNode* dummy = new ListNode();
        ListNode* tail = dummy;
        int s = 0;
        for (ListNode* cur = head->next; cur; cur = cur->next) {
            if (cur->val)
                s += cur->val;
            else {
                tail->next = new ListNode(s);
                tail = tail->next;
                s = 0;
            }
        }
        return dummy->next;
    }
};

/**
 * Definition for singly-linked list.
 * type ListNode struct {
 *     Val int
 *     Next *ListNode
 * }
 */
func mergeNodes(head *ListNode) *ListNode {
    dummy := &ListNode{}
    tail := dummy
    s := 0
    for cur := head.Next; cur != nil; cur = cur.Next {
        if cur.Val != 0 {
            s += cur.Val
        } else {
            tail.Next = &ListNode{Val: s}
            tail = tail.Next
            s = 0
        }
    }
    return dummy.Next
}

/**
 * Definition for singly-linked list.
 * class ListNode {
 *     val: number
 *     next: ListNode | null
 *     constructor(val?: number, next?: ListNode | null) {
 *         this.val = (val===undefined ? 0 : val)
 *         this.next = (next===undefined ? null : next)
 *     }
 * }
 */

function mergeNodes(head: ListNode | null): ListNode | null {
    const dummy = new ListNode();
    let cur = dummy;
    let sum = 0;
    while (head) {
        if (head.val === 0 && sum !== 0) {
            cur.next = new ListNode(sum);
            cur = cur.next;
            sum = 0;
        }
        sum += head.val;
        head = head.next;
    }
    return dummy.next;
}

// Definition for singly-linked list.
// #[derive(PartialEq, Eq, Clone, Debug)]
// pub struct ListNode {
//   pub val: i32,
//   pub next: Option<Box<ListNode>>
// }
//
// impl ListNode {
//   #[inline]
//   fn new(val: i32) -> Self {
//     ListNode {
//       next: None,
//       val
//     }
//   }
// }
impl Solution {
    pub fn merge_nodes(mut head: Option<Box<ListNode>>) -> Option<Box<ListNode>> {
        let mut dummy = Box::new(ListNode::new(-1));
        let mut cur = &mut dummy;
        let mut sum = 0;
        while let Some(node) = head {
            if node.val == 0 && sum != 0 {
                cur.next = Some(Box::new(ListNode::new(sum)));
                cur = cur.as_mut().next.as_mut().unwrap();
                sum = 0;
            }
            sum += node.val;
            head = node.next;
        }
        dummy.next.take()
    }
}

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     struct ListNode *next;
 * };
 */

struct ListNode* mergeNodes(struct ListNode* head) {
    struct ListNode dummy;
    struct ListNode* cur = &dummy;
    int sum = 0;
    while (head) {
        if (head->val == 0 && sum != 0) {
            cur->next = malloc(sizeof(struct ListNode));
            cur->next->val = sum;
            cur->next->next = NULL;
            cur = cur->next;
            sum = 0;
        }
        sum += head->val;
        head = head->next;
    }
    return dummy.next;
}

2181. 合并零之间的节点

题目描述

解法

方法一

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