Question

2181. Merge Nodes in Between Zeros

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Description

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.

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.

Return _the_ head _of the modified linked list_.

 

Example 1:

Input: head = [0,3,1,0,4,5,2,0]
Output: [4,11]
Explanation: 
The above figure represents the given linked list. The modified list contains
- The sum of the nodes marked in green: 3 + 1 = 4.
- The sum of the nodes marked in red: 4 + 5 + 2 = 11.

Example 2:

Input: head = [0,1,0,3,0,2,2,0]
Output: [1,3,4]
Explanation: 
The above figure represents the given linked list. The modified list contains
- The sum of the nodes marked in green: 1 = 1.
- The sum of the nodes marked in red: 3 = 3.
- The sum of the nodes marked in yellow: 2 + 2 = 4.

 

Constraints:

• The number of nodes in the list is in the range [3, 2 * 105].
• 0 <= Node.val <= 1000
• There are no two consecutive nodes with Node.val == 0.
• The beginning and end of the linked list have Node.val == 0.

Solutions

Solution 1

# 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;
}