> For the complete documentation index, see [llms.txt](https://hao-fu-1.gitbook.io/oj/llms.txt). Markdown versions of documentation pages are available by appending `.md` to page URLs; this page is available as [Markdown](https://hao-fu-1.gitbook.io/oj/stack/wei-chi-di-zeng-huo-di-jian/next-greater-element-i/next-greater-node-in-linked-list.md).

# Next Greater Node In Linked List

<https://leetcode.com/contest/weekly-contest-130/problems/next-greater-node-in-linked-list/>

> 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]

## Solution

和next greater number类似，用一个stack，如果递减就存入，直到遇到第一个不满足的元素，此时pop直到满足为止，pop出的元素它们的右边第一大即该元素。由于是list，不知道抛出元素所在位置，因此stack 与其存element值，不如存element所在位置。对应的值先存在res里，然后再被更新。栈顶对应的值即res\[stack.back()].

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

        while (!s.empty()) {
            res[s.top()] = 0;
            s.pop();
        }
        return res;
    }
};
```