1019 链表的下一个更大节点（单调栈）

题目

链表的下一个更大节点

给定一个长度为 n 的链表 head

对于列表中的每个节点，查找下一个 更大节点 的值。也就是说，对于每个节点，找到它旁边的第一个节点的值，这个节点的值 严格大于 它的值。

返回一个整数数组 answer ，其中 answer[i] 是第 i 个节点( 从1开始 )的下一个更大的节点的值。如果第 i 个节点没有下一个更大的节点，设置 answer[i] = 0 。

示例 1：

输入：head = [2,1,5]
输出：[5,5,0]

示例 2：

输入：head = [2,7,4,3,5]
输出：[7,0,5,5,0]

提示：

• 链表中节点数为 n
• 1 <= n <= 104
• 1 <= Node.val <= 109

题解

从右向左

寻找下一个更大元素

/**
 * 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 {
    private int n;
 
    public int[] nextLargerNodes(ListNode head) {
        //从右向左 查找下一个更大元素
        head = reverse(head);
        int[] ans = new int[n];
        Deque st = new ArrayDeque<>();
        for (ListNode cur = head; cur != null; cur = cur.next) {
            while (!st.isEmpty() && st.peek() <= cur.val) {
                st.pop();
            }
            ans[--n] = st.isEmpty() ? 0 : st.peek();
            st.push(cur.val);
        }
        return ans;
    }
 
    public ListNode reverse(ListNode head) {
        ListNode pre = null;
        ListNode cur = head;
        while (cur != null) {
            ListNode nxt = cur.next;
            cur.next = pre;
            pre = cur;
            cur = nxt;
            n++;
        }
        return pre;
    }
}

也可以使用递归倒序遍历

/**
 * 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 {
    private int[] ans;
    private final Deque st = new ArrayDeque<>();
 
    public int[] nextLargerNodes(ListNode head) {
        f(head, 0);
        return ans;
    }
 
    private void f(ListNode head, int i) {
        if (head == null) {
            ans = new int[i];
            return;
        }
        f(head.next, i + 1);
        while (!st.isEmpty() && st.peek() <= head.val) {
            st.pop();
        }
        if (!st.isEmpty()) {
            ans[i] = st.peek();
        }
        st.push(head.val);
    }
}

从左向右

更新下一个更大元素

/**
 * 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) {
        int n = 0;
        for (ListNode cur = head; cur != null; cur = cur.next) {
            n++;
        }
        int[] ans = new int[n];
        Deque<int[]> st = new ArrayDeque<>();// 保存节点值和节点下标
        int i = 0;
        for (ListNode cur = head; cur != null; cur = cur.next) {
            while (!st.isEmpty() && st.peek()[0] < cur.val) {
                ans[st.pop()[1]] = cur.val;// 更新答案
            }
            st.push(new int[]{cur.val, i++});
        }
        return ans;
    }
}

/**
 * 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) {
        int n = 0;
        for (ListNode cur = head; cur != null; cur = cur.next) {
            n++;
        }
        int[] ans = new int[n];
        Deque st = new ArrayDeque<>();// 保存节点下标
        int i = 0;
        for (ListNode cur = head; cur != null; cur = cur.next) {
            while (!st.isEmpty() && ans[st.peek()] < cur.val) {
                ans[st.pop()] = cur.val;// 更新答案
            }
            st.push(i);
            ans[i++] = cur.val;
        }
        for (var index : st) {
            ans[index] = 0;
        }
        return ans;
    }
}