从尾到头打印链表输入一个链表的头节点, 从尾到头反过来返回每个节点的值(用数组返回). <div class="snipp

解题思路使用栈(双端队列), 先将链表中的值推入栈中, 再从栈顶依次添加到数组中. <h3 di

LeetCode题解解题思路使用栈将链表元素顺序倒置. 从

从尾到头打印链表 about pisecespeng.record.me HOT 2 CLOSED

PisecesPeng commented on September 7, 2024

从尾到头打印链表

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Comments (2)

PisecesPeng commented on September 7, 2024

解题思路

使用栈(双端队列), 先将链表中的值推入栈中, 再从栈顶依次添加到数组中.

代码

class ListNode {
    int val;
    ListNode next;
    ListNode(int x) {
        val = x;
    }
}

public static int[] func(ListNode head) {
    if (head == null)
        return new int[0];

    ArrayDeque<Integer> arrayDeque = new ArrayDeque();
    do {
        arrayDeque.addFirst(head.val);
        head = head.next;
    } while (head != null);

    int[] r = new int[arrayDeque.size()];
    for (int i = 0, size = arrayDeque.size(); i < size; i++)
        r[i] = arrayDeque.pollFirst();
    return r;
}

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PisecesPeng commented on September 7, 2024

LeetCode题解

解题思路

使用栈将链表元素顺序倒置.
从链表的头节点开始，依次将每个节点压入栈内，然后依次弹出栈内的元素并存储到数组中

代码

class ListNode {
    int val;
    ListNode next;
    ListNode(int x) {
        val = x;
    }
}

public static int[] func(ListNode head) {
    Stack<ListNode> stack = new Stack<ListNode>();
    ListNode temp = head;
    while (temp != null) {
        stack.push(temp);
        temp = temp.next;
    }
    int size = stack.size();
    int[] print = new int[size];
    for (int i = 0; i < size; i++) {
        print[i] = stack.pop().val;
    }
    return print;
}