LeetCode每日一题(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.

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这题应该会有很多种解法，遍历+递归或者单纯的遍历都可以解决问题， 或者单纯的递归应该也可以， 只是写法上可能要复杂一些。我选的第一种方式， 遍历+递归， 遇到 0 则进行递归调用得到后面的链表， 如果不是 0 则继续向后遍历累加当前 node 的 val

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impl Solution {
    pub fn merge_nodes(mut head: Option<Box<ListNode>>) -> Option<Box<ListNode>> {
        let mut curr = 0;
        while let Some(mut node) = head {
            if node.val == 0 {
                if curr == 0 {
                    head = node.next.take();
                    continue;
                }
                let mut n = ListNode::new(curr);
                n.next = Solution::merge_nodes(node.next.take());
                return Some(Box::new(n));
            }
            head = node.next.take();
            curr += node.val;
        }
        None
    }
}

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