1019. Next Greater Node in Linked List

1019. Next Greater Node In Linked List #

题目 #

We are given a linked list with head as the first node. Let’s number the nodes in the list: node_1, node_2, node_3, … etc.

Each node may have a next larger value: for node_i, next_larger(node_i) is the node_j.val such that j > i, node_j.val > node_i.val, and j is the smallest possible choice. If such a j does not exist, the next larger value is 0.

Return an array of integers answer, where answer[i] = next_larger(node_{i+1}).

Note that in the example inputs (not outputs) below, arrays such as [2,1,5] represent the serialization of a linked list with a head node value of 2, second node value of 1, and third node value of 5.

Example 1:


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

Example 2:


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

Example 3:


Input: [1,7,5,1,9,2,5,1]
Output: [7,9,9,9,0,5,0,0]

Note:

  • 1 <= node.val <= 10^9 for each node in the linked list.
  • The given list has length in the range [0, 10000].

题目大意 #

给出一个链表,要求找出每个结点后面比该结点值大的第一个结点,如果找不到这个结点,则输出 0 。

解题思路 #

这一题和第 739 题、第 496 题、第 503 题类似。也有 2 种解题方法。先把链表中的数字存到数组中,整道题的思路就和第 739 题完全一致了。普通做法就是 2 层循环。优化的做法就是用单调栈,维护一个单调递减的栈即可。

代码 #


package leetcode

/**
 * Definition for singly-linked list.
 * type ListNode struct {
 *     Val int
 *     Next *ListNode
 * }
 */
// 解法一 单调栈
func nextLargerNodes(head *ListNode) []int {
	res, indexes, nums := make([]int, 0), make([]int, 0), make([]int, 0)
	p := head
	for p != nil {
		nums = append(nums, p.Val)
		p = p.Next
	}
	for i := 0; i < len(nums); i++ {
		res = append(res, 0)
	}
	for i := 0; i < len(nums); i++ {
		num := nums[i]
		for len(indexes) > 0 && nums[indexes[len(indexes)-1]] < num {
			index := indexes[len(indexes)-1]
			res[index] = num
			indexes = indexes[:len(indexes)-1]
		}
		indexes = append(indexes, i)
	}
	return res
}


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