LeetCode 1019: Next Greater Node In Linked List Solution

Master LeetCode problem 1019 (Next Greater Node In Linked List), a medium challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

1019. Next Greater Node In Linked List

Medium

Array Linked List Stack Monotonic Stack

Problem Explanation

Explanation

To solve this problem, we can iterate through the linked list and maintain a stack to store the indices of nodes whose next greater node is yet to be found. We traverse the list in reverse order to find the next greater node for each node. We push the index of the current node onto the stack and keep popping indices from the stack until the top of the stack represents a node with a value greater than the current node. For each popped index, we update the corresponding answer in the result array. Finally, we update the current node's next greater node index by pushing the current index onto the stack.

Time Complexity: O(N)
Space Complexity: O(N)

Solution Code

class Solution {
    public int[] nextLargerNodes(ListNode head) {
        List<Integer> list = new ArrayList<>();
        while (head != null) {
            list.add(head.val);
            head = head.next;
        }
        
        int n = list.size();
        int[] result = new int[n];
        Stack<Integer> stack = new Stack<>();
        
        for (int i = n - 1; i >= 0; i--) {
            while (!stack.isEmpty() && list.get(i) >= list.get(stack.peek())) {
                stack.pop();
            }
            result[i] = stack.isEmpty() ? 0 : list.get(stack.peek());
            stack.push(i);
        }
        
        return result;
    }
}

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