1019. Next Greater Node In Linked List

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

题意：给定一个带有头节点的单链表，对链表中的每个元素，求出右面第一个大于它的值，不存在记为0。

思路：先将链表中的数据存到动态数组中，便于操作。因为要求的是右面的大值，我们可以从后向前遍历，维护一个单调递减栈。若小于栈顶元素时，则栈顶元素记为大值；否则，栈中元素出栈，直至找到大值。

class Solution {
public:
    vector<int> nextLargerNodes(ListNode* head) {
        vector<int>vec,sc;
        stack<int>st;
        while(head){
             vec.push_back(head->val);
             sc.push_back(0);
             head=head->next;
        }
        st.push(vec[vec.size()-1]);
        for(int i=vec.size()-2;i>=0;i--){
            while(!st.empty()&&vec[i]>=st.top())
                     st.pop();
            if(!st.empty())
                 sc[i]=st.top();
            st.push(vec[i]);
        }
        return sc;
    }
};