1019. Next Greater Node In Linked List**
https://leetcode.com/problems/next-greater-node-in-linked-list/
题目描述
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]
.
C++ 实现 1
可以考虑先看看 496. Next Greater Element I* 这道题找找感觉. 使用栈来做比较方便.
先用 record
将链表中的每一个数保存下来, 然后依次访问 record
, 查找每个元素的 NGE. 栈中保存着那些暂时还没有找到 NGE 的元素的索引.
class Solution {
public:
vector<int> nextLargerNodes(ListNode* head) {
stack<int> st;
vector<int> record;
auto p = head;
while (p) {
record.push_back(p->val);
p = p->next;
}
vector<int> res(record.size(), 0);
for (int i = 0; i < record.size(); ++ i) {
while (!st.empty() && record[i] > record[st.top()]) {
res[st.top()] = record[i];
st.pop();
}
st.push(i);
}
return res;
}
};
C++ 实现 2
栈 st
中保存目前还没有找到 NGE 的节点, record
记录每个节点对应的 NGE 的值.
class Solution {
public:
vector<int> nextLargerNodes(ListNode* head) {
stack<ListNode*> st;
vector<int> res;
unordered_map<ListNode*, int> record;
auto p = head;
while (p) {
while (!st.empty() && p->val > st.top()->val) {
record[st.top()] = p->val;
st.pop();
}
st.push(p);
p = p->next;
}
p = head;
while (p) {
if (record.count(p)) res.push_back(record[p]);
else res.push_back(0);
p = p->next;
}
return res;
}
};