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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.

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.

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.

풀이

재영짱짱1234


/**
 * Definition for singly-linked list.
 * function ListNode(val, next) {
 *     this.val = (val===undefined ? 0 : val)
 *     this.next = (next===undefined ? null : next)
 * }
 */
/**
 * @param {ListNode} head
 * @return {ListNode}
 */
var mergeNodes = function(head) {
    let cacheHead = head;
		
		// 맨처음부터 0 0 0 4 5  -> 4 5 부터 시작하도록
    while (!cacheHead.val) {
        cacheHead = cacheHead.next
    }
    
    let prevCacheNode; // 이전의 노드를 캐시.
    
		// merge node - 0일 때 -> 연결해주고 + 다음에 숫자가 오는 노드를 캐시
    while (head) {
        if (!head.val) {
            if (prevCacheNode) {
                prevCacheNode.next = head.next; // 연결의 의미
            }
            
            prevCacheNode = head.next; // 다음 캐시된 노드를 설정함. head.next - 숫자가 올 때까지 설정해야 함.
            head = head.next;
            
            continue;
        }
        
        if (head.next) {
            prevCacheNode.val += head.next.val; .// 이전 캐시 노드는 val !== 0 -> +
        }
        
        head = head.next;
    }
    
    return cacheHead;
};

은찬


var mergeNodes = function(head) {
    const answer = new ListNode();
    const arr =[];
    let current = head;
    let sum = 0;
    
    while(current !== null){
        if(current.val !== 0){
            sum += current.val;
        } else{
            if(sum !== 0){
                arr.push(sum);
                sum = 0;
            }
        }
        current = current.next;
    }
    
    current = answer;
    
    for(let i = 0; i < arr.length; i++){
        current.val = arr[i];
        
        if(i < arr.length - 1){
            current.next = new ListNode();
            current = current.next;
        }
    }
    
    return answer;
};

효성님을 위해 미리 만들어놓았습니다ㅋㅋㅋ

(감사합니다 ㅋㅋ)


var mergeNodes = function(head) {
    const mockHead = new ListNode();
    let mergedNode = mockHead;
    let cur = head;
    
    while(cur) {
        if(cur.val === 0 && cur.next) {
            mergedNode.next = new ListNode();
            mergedNode = mergedNode.next;
        }
        mergedNode.val += cur.val;
        cur = cur.next;
    }
    
    return mockHead.next;
};

Merge Nodes in Between Zeros

풀이