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1019. Next Greater Node In Linked List

中文文档

Description

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:

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

Example 2:

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

Solutions

Python3

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, x):
#         self.val = x
#         self.next = None

class Solution:
    def nextLargerNodes(self, head: ListNode) -> List[int]:
        nums = []
        while head:
            nums.append(head.val)
            head = head.next
        s = []
        larger = [0] * len(nums)
        for i, num in enumerate(nums):
            while s and nums[s[-1]] < num:
                larger[s.pop()] = num
            s.append(i)
        return larger

Java

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
class Solution {
    public int[] nextLargerNodes(ListNode head) {
        List<Integer> nums = new ArrayList<>();
        while (head != null) {
            nums.add(head.val);
            head = head.next;
        }
        Deque<Integer> s = new ArrayDeque<>();
        int[] larger = new int[nums.size()];
        for (int i = 0; i < nums.size(); ++i) {
            while (!s.isEmpty() && nums.get(s.peek()) < nums.get(i)) {
                larger[s.pop()] = nums.get(i);
            }
            s.push(i);
        }
        return larger;
    }
}

JavaScript

/**
 * Definition for singly-linked list.
 * function ListNode(val) {
 *     this.val = val;
 *     this.next = null;
 * }
 */
/**
 * @param {ListNode} head
 * @return {number[]}
 */
var nextLargerNodes = function (head) {
    let nums = [];
    while (head != null) {
        nums.push(head.val);
        head = head.next;
    }
    const n = nums.length;
    let larger = new Array(n).fill(0);
    let stack = [];
    for (let i = 0; i < n; i++) {
        let num = nums[i];
        while (stack.length > 0 && nums[stack[stack.length - 1]] < num) {
            larger[stack.pop()] = num;
        }
        stack.push(i);
    }
    return larger;
};

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