2487. Remove Nodes From Linked List-优快云博客

本文介绍了一种算法，给定一个单向链表，需要移除所有右侧存在较大值节点的链表节点。使用栈实现了一个迭代方法，通过遍历链表并检查节点值来完成操作。

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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.

Example 1:

Input: head = [5,2,13,3,8]
Output: [13,8]
Explanation: The nodes that should be removed are 5, 2 and 3.
- Node 13 is to the right of node 5.
- Node 13 is to the right of node 2.
- Node 8 is to the right of node 3.

Example 2:

Input: head = [1,1,1,1]
Output: [1,1,1,1]
Explanation: Every node has value 1, so no nodes are removed.

/**
 * 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 *> st;
        while(head->next != nullptr){
            st.push(head);
            head = head->next;
        }
        while(!st.empty()){
            if(st.top()->val >=head->val||head==nullptr){
                st.top()->next = head;
                head = st.top();
            }
            st.pop();
        }
        return head;

    }
};