Summary of Ripple Solution 1019. Next Larger Node in Linked List

Directory link:

Lituo Programming Problems - Summary of Solutions_Share+Records-CSDN Blog

GitHub synchronous brushing project:

https://github.com/September26/java-algorithms

Link to the original title: Likou

describe:

Given a  n linked list  of lengthhead

For each node in the list, find the value of the next  greater node  . That is, for each node, find the value of the first node next to it that is  strictly greater than  its value.

Returns an array of integers  answer , where  answer[i] is   the value of the next greater node of the -th i node (  starting from 1 ). Set if the  i th node has no next greater node  answer[i] = 0 .

Example 1:

 

Input: head = [2,1,5]
 Output: [5,5,0]

Example 2:

 

Input: head = [2,7,4,3,5]
 Output: [7,0,5,5,0]

hint:

• The number of nodes in the linked list is n
• 1 <= n <= 104
• 1 <= Node.val <= 109

Problem-solving ideas:

* Problem-solving ideas:
* Build a stack structure stack and traverse from back to front.
* If the current value >= the value at the top of the stack, pop this value from the top of the stack. Continue to take the value at the top of the stack for comparison.
* If the stack is empty, it means that there is no larger value behind, then set the value bit of the current position to 0.
* If the current value < the value at the top of the stack, it means that the greater value at the current position is the value at the top of the stack.

code:

public class Solution1019 {

    public int[] nextLargerNodes(ListNode head) {
        Stack<Integer> stack = new Stack<>();
        List<Integer> list = new ArrayList<>();
        while (head != null) {
            list.add(head.val);
            head = head.next;
        }
        int[] arr = new int[list.size()];
        for (int i = list.size() - 1; i >= 0; i--) {
            Integer value = list.get(i);
            while (stack.size() > 0) {
                Integer peek = stack.peek();
                if (value >= peek) {
                    stack.pop();
                } else {
                    break;
                }
            }
            if (stack.size() == 0) {
                arr[i] = 0;
            } else {
                arr[i] = stack.peek();
            }
            stack.add(value);
        }
        return arr;
    }
}