1019. Next Greater Node In Linked List

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1019. Next Greater Node In Linked List

Description

Difficulty: Medium

Related Topics: Array, Linked List, Stack, Monotonic Stack

You are given the head of a linked list with n nodes.

For each node in the list, find the value of the next greater node. That is, for each node, find the value of the first node that is next to it and has a strictly larger value than it.

Return an integer array answer where answer[i] is the value of the next greater node of the ith node (1-indexed). If the ith node does not have a next greater node, set answer[i] = 0.

Example 1:

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Input: head = [2,1,5]
Output: [5,5,0]

Example 2:

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Input: head = [2,7,4,3,5]
Output: [7,0,5,5,0]

Constraints:

  • The number of nodes in the list is n.
  • 1 <= n <= 104
  • 1 <= Node.val <= 109

Hints/Notes

  • transform the linked list to array first
  • stack

Solution

Language: C++

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/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode() : val(0), next(nullptr) {}
 *     ListNode(int x) : val(x), next(nullptr) {}
 *     ListNode(int x, ListNode *next) : val(x), next(next) {}
 * };
 */
class Solution {
public:
    vector<int> nextLargerNodes(ListNode* head) {
        vector<int> nums;
        while (head) {
            nums.push_back(head->val);
            head = head->next;
        }
        stack<int> s;
        vector<int> res(nums.size(), 0);
        for (int i = nums.size() - 1; i >= 0; i--) {
            while (!s.empty() && (s.top() <= nums[i])) {
                s.pop();
            }
            res[i] = s.empty() ? 0 : s.top();
            s.push(nums[i]);
        }
        return res;
    }
};