Leetcode 1019 Next Greater Node

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.

简而言之：找下一个比node_i大的值node_j，且j最小。如果没有则置0

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. 1 <= node.val <= 10^9 for each node in the linked list.
2. The given list has length in the range [0, 10000].

解题思路

很简单，一开始想复杂了。。。

只需要找到下一个比node_i大的值即可。两个循环，外层循环标记当前节点node_i，内层循环遍历node_i之后的结点，只要找到第一个比node_i大的结点就退出内层循环，然后给结果数组赋值。时间复杂度O(n^2)。

题解

    public static int[] nextLargerNodes(ListNode head) {
        // 异常处置部分
        if (head == null ) return null;
        int n = 0;
        ListNode p = head;
        while(p != null){
            p = p.next;
            n++;
        }
        if (n == 0 ) return null;
        // 以上
        int []result = new int[n];
        ListNode tail = head;
        // i来标记返回数组的位置，以tail是同步的
        int i = 0;
        // tail 为 node_i
        while(tail != null){
            int max = 0;
            p = tail;
            // p 为 node_j
            while(p != null){
                if (p.val > tail.val){
                    max = p.val;
                    break;
                }
                p=p.next;
            }
            // 如果存在较大值 则max就是，不存在 则max默认为0；
            result[i] = max;
            i++;
            tail = tail.next;
        }
        return result;
    }

大神解法

借助栈来实现，时间复杂度O(n),空间复杂度O(n)。

思路：栈内存放的是 结点的编号。 每压入一个编号前，先检查该编号对应元素是否大于栈顶的编号对应元素，、如果大于，则栈顶编号出列，且栈顶编号对应元素的greater为 当前编号对应元素。如果不大于，则继续压栈。

使用ArrayList方便获取编号和元素的对应关系，且无需固定长度数组。

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