LC.P1019[链表中的下一个更大节点]

思路

利用单调栈，从右到左遍历链表的数据，若栈顶元素$<=$当前遍历到的元素，则弹出栈。

若栈中仍有元素，则存在下一个更大的节点；否则为0。

方法一：单调栈(List实现反转)

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/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode() {}
 *     ListNode(int val) { this.val = val; }
 *     ListNode(int val, ListNode next) { this.val = val; this.next = next; }
 * }
 */
class Solution {
    public int[] nextLargerNodes(ListNode head) {
        List<Integer> list = new ArrayList<>();
        while (head != null) {
            list.add(head.val);
            head = head.next;
        }
        int n = list.size();
        int[] ans = new int[n];
        Deque<Integer> stack = new ArrayDeque<>();
        for (int i = n - 1; i >= 0; --i) {
            int num = list.get(i);
            while (!stack.isEmpty() && stack.peek() <= num) stack.pop();
            if (!stack.isEmpty()) ans[i] = stack.peek();
            stack.push(num);
        }
        return ans;
    }
}
  • 时间复杂度：$O(n)$
  • 空间复杂度：$O(n)$

方法二：单调栈(反转链表)

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/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode() {}
 *     ListNode(int val) { this.val = val; }
 *     ListNode(int val, ListNode next) { this.val = val; this.next = next; }
 * }
 */
class Solution {
    int n;

    public int[] nextLargerNodes(ListNode head) {
        head = reverse(head);
        int[] ans = new int[n];
        Deque<Integer> stack = new ArrayDeque<>();
        while (head != null) {
            int num = head.val;
            --n;
            while (!stack.isEmpty() && stack.peek() <= num) stack.pop();
            if (!stack.isEmpty()) ans[n] = stack.peek();
            stack.push(num);
            head = head.next;
        }
        return ans;
    }

    /**
     * 反转链表
     */
    private ListNode reverse(ListNode head) {
        ListNode pre = null, cur = head;
        while (cur != null) {
            ListNode temp = cur.next;
            cur.next = pre;
            pre = cur;
            cur = temp;
            ++n;
        }
        return pre;
    }
}
  • 时间复杂度：$O(n)$
  • 空间复杂度：$O(n)$

方法三：单调栈(递归实现反转)

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/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode() {}
 *     ListNode(int val) { this.val = val; }
 *     ListNode(int val, ListNode next) { this.val = val; this.next = next; }
 * }
 */
class Solution {
    int[] ans;
    Deque<Integer> stack;

    public int[] nextLargerNodes(ListNode head) {
        stack = new ArrayDeque<>();
        func(head, 0);
        return ans;
    }

    /**
     * 构造答案
     *
     * @param node 当前节点
     * @param n    链表长度
     */
    private void func(ListNode node, int n) {
        if (node == null) {
            ans = new int[n];
            return;
        }
        func(node.next, n + 1);
        while (!stack.isEmpty() && stack.peek() <= node.val) stack.pop();
        if (!stack.isEmpty()) ans[n] = stack.peek();
        stack.push(node.val);
    }
}
  • 时间复杂度：$O(n)$
  • 空间复杂度：$O(n)$