暴力破解与最直观的写法：

/**
 * 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> list;
        while(head)
        {
            list.push_back(head->val);
            head = head->next;
        }

        int size = list.size();
        for(int i = 0,j = 0;i < size;i++)
        {
            bool flag = false;
            for(j = i + 1;j < size;j++)
            {
                if(list[j] > list[i])
                {
                    list[i] = list[j];
                    flag = true;
                    break;
                }       
            }
            if(j == size && flag == false)
                list[i] = 0;
        }
        list[size - 1] = 0;
        return list;
    }
};

毫无意外的执行结果超时：

Time Limit Exceeded
Details 
Last executed input
[10000,10000,10000,10000,10000,10000,9999,9998,9998,9996,9994,9992,9992,9990,9990,9989,……]

优化后的代码如下：

/**
 * 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> res,stack;
        
        while(head)
        {
            res.push_back(head->val);
            head = head->next;
        }
        for(int i = res.size() - 1;i >= 0;i--)
        {
            auto val = res[i];
            while(!stack.empty() && val >= stack.back()) stack.pop_back();
            
            res[i] = stack.empty() ? 0:stack.back();
            stack.push_back(val);
           
        }
        
        return res;
    }
};

这里从后往前处理，初始时stack为空，res最后元素为0，然后将修改前的val存入stack,后面判断stack只要非空，判断当前值只要不比stack里的元素小，就讲元素弹出，这样就保留了最新的最大值。
程序执行结果：

Runtime: 144 ms, faster than 40.57% of C++ online submissions for Next Greater Node In Linked List.
Memory Usage: 41.8 MB, less than 57.03% of C++ online submissions for Next Greater Node In Linked List.