[LeetCode] 2181. Merge Nodes in Between Zeros

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.

合并零之间的节点。

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 。

思路

这是链表类的简单题，请直接参考代码注释。

复杂度

时间O(n)
空间O(1)

代码

Java实现

/**
 * 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(0);
        ListNode p = dummy;
        ListNode cur = head;
        int sum = 0;
        while (cur != null) {
			// 遇到0就开始计算0之间的node的和
            if (cur.val == 0) {
                sum = 0;
                cur = cur.next;
                while (cur != null && cur.val != 0) {
                    sum += cur.val;
                    cur = cur.next;
                }
				// 遇到下一个0或者list结尾就把之前计算的和放到新的node中
                if (sum != 0) {
                    p.next = new ListNode(sum);
                    p = p.next;
                }
            } else {
                cur = cur.next;
            }
        }
        return dummy.next;
    }
}