1030.链表中下一个最大的节点

这道题就是找自己大的那个数就好 简单的解法

class Solution {
    public int[] nextLargerNodes(ListNode head) {
        int arr[] = new int[100001];
        int flag = 0;
        while (head != null) {
            arr[flag++] = head.val;
            head = head.next;
        }
        int res[] = new int[flag];
        for (int i = 0; i < flag; i++) {
            for (int k = i + 1; k < flag; k++) {
                if (arr[k] > arr[i]) {
                    res[i] = arr[k];
                    break;
                }
            }
        }
        return res;
    }
}

后面增加用栈解的方法

class Solution {
    public int[] nextLargerNodes(ListNode head) {
        ArrayList<Integer> al = new ArrayList<Integer>();
        while(head!=null)
        {al.add(head.val);
            head = head.next;
        }
        int res[] = new int [al.size()];
        Stack<Integer> s = new Stack<Integer>();
        for (int i = 0; i < al.size(); ++i) {
            while (!s.isEmpty() && al.get(s.peek()) < al.get(i))
                res[s.pop()] = al.get(i);
            s.push(i);
        }
        return res;
    }
}

1030.链表中下一个最大的节点

原文：https://www.cnblogs.com/cznczai/p/11150371.html