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:
[3, 2 * 105].0 <= Node.val <= 1000Node.val == 0.Node.val == 0.
class Solution {
public ListNode mergeNodes(ListNode head) {
ListNode dummy = new ListNode(-1);
ListNode ans = dummy ;
ListNode curr = head.next;
int sum = 0 ;
while(curr!=null){
if(curr.val !=0){
sum += curr.val;
}else{
ListNode temp = new ListNode(sum);
dummy.next = temp ;
dummy = dummy.next;
sum = 0 ;
}
curr = curr.next;
}
return ans.next;
}
}
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