Description
You are given the head of a linked list.
Remove every node which has a node with a greater value anywhere to the right side of it.
Return the head of the modified linked list.
Solution
Since every node needs to have a node with a greater value anywhere to the right side of it, the output linked list must be a decreasing sequence.
Therefore, to solve this problem, we will need to reverse the linked list using a stack and form the new list by checking the values when popping out the nodes.
Implementation
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/**
* 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:
ListNode* removeNodes(ListNode* head) {
stack<ListNode*> nodes;
ListNode* cur = head;
ListNode* newHead = nullptr;
while (cur) {
nodes.push(cur);
cur = cur->next;
}
int max = 0;
while (!nodes.empty()) {
ListNode* node = nodes.top();
nodes.pop();
if (node->val >= max) {
max = node->val;
node->next = newHead;
newHead = node;
}
else {
delete node;
}
}
return newHead;
}
};