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2816. 翻倍以链表形式表示的数字

题目描述

给你一个 非空 链表的头节点 head ，表示一个不含前导零的非负数整数。

将链表 翻倍 后，返回头节点 head 。

 

示例 1：

输入：head = [1,8,9]
输出：[3,7,8]
解释：上图中给出的链表，表示数字 189 。返回的链表表示数字 189 * 2 = 378 。

示例 2：

输入：head = [9,9,9]
输出：[1,9,9,8]
解释：上图中给出的链表，表示数字 999 。返回的链表表示数字 999 * 2 = 1998 。

 

提示：

  • 链表中节点的数目在范围 [1, 104] 内
  • 0 <= Node.val <= 9
  • 生成的输入满足：链表表示一个不含前导零的数字，除了数字 0 本身。

解法

方法一：翻转链表 + 模拟

我们先将链表翻转，然后模拟乘法运算，最后再将链表翻转回来。

时间复杂度 $O(n)$，其中 $n$ 是链表的长度。忽略答案链表的空间消耗，空间复杂度 $O(1)$。

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# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
class Solution:
    def doubleIt(self, head: Optional[ListNode]) -> Optional[ListNode]:
        def reverse(head):
            dummy = ListNode()
            cur = head
            while cur:
                next = cur.next
                cur.next = dummy.next
                dummy.next = cur
                cur = next
            return dummy.next

        head = reverse(head)
        dummy = cur = ListNode()
        mul, carry = 2, 0
        while head:
            x = head.val * mul + carry
            carry = x // 10
            cur.next = ListNode(x % 10)
            cur = cur.next
            head = head.next
        if carry:
            cur.next = ListNode(carry)
        return reverse(dummy.next)

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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 ListNode doubleIt(ListNode head) {
        head = reverse(head);
        ListNode dummy = new ListNode();
        ListNode cur = dummy;
        int mul = 2, carry = 0;
        while (head != null) {
            int x = head.val * mul + carry;
            carry = x / 10;
            cur.next = new ListNode(x % 10);
            cur = cur.next;
            head = head.next;
        }
        if (carry > 0) {
            cur.next = new ListNode(carry);
        }
        return reverse(dummy.next);
    }

    private ListNode reverse(ListNode head) {
        ListNode dummy = new ListNode();
        ListNode cur = head;
        while (cur != null) {
            ListNode next = cur.next;
            cur.next = dummy.next;
            dummy.next = cur;
            cur = next;
        }
        return dummy.next;
    }
}

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/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode() : val(0), next(nullptr) {}
 *     ListNode(int x) : val(x), next(nullptr) {}
 *     ListNode(int x, ListNode *next) : val(x), next(next) {}
 * };
 */
class Solution {
public:
    ListNode* doubleIt(ListNode* head) {
        head = reverse(head);
        ListNode* dummy = new ListNode();
        ListNode* cur = dummy;
        int mul = 2, carry = 0;
        while (head) {
            int x = head->val * mul + carry;
            carry = x / 10;
            cur->next = new ListNode(x % 10);
            cur = cur->next;
            head = head->next;
        }
        if (carry) {
            cur->next = new ListNode(carry);
        }
        return reverse(dummy->next);
    }

    ListNode* reverse(ListNode* head) {
        ListNode* dummy = new ListNode();
        ListNode* cur = head;
        while (cur) {
            ListNode* next = cur->next;
            cur->next = dummy->next;
            dummy->next = cur;
            cur = next;
        }
        return dummy->next;
    }
};

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/**
 * Definition for singly-linked list.
 * type ListNode struct {
 *     Val int
 *     Next *ListNode
 * }
 */
func doubleIt(head *ListNode) *ListNode {
    head = reverse(head)
    dummy := &ListNode{}
    cur := dummy
    mul, carry := 2, 0
    for head != nil {
        x := head.Val*mul + carry
        carry = x / 10
        cur.Next = &ListNode{Val: x % 10}
        cur = cur.Next
        head = head.Next
    }
    if carry > 0 {
        cur.Next = &ListNode{Val: carry}
    }
    return reverse(dummy.Next)
}

func reverse(head *ListNode) *ListNode {
    dummy := &ListNode{}
    cur := head
    for cur != nil {
        next := cur.Next
        cur.Next = dummy.Next
        dummy.Next = cur
        cur = next
    }
    return dummy.Next
}

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/**
 * Definition for singly-linked list.
 * class ListNode {
 *     val: number
 *     next: ListNode | null
 *     constructor(val?: number, next?: ListNode | null) {
 *         this.val = (val===undefined ? 0 : val)
 *         this.next = (next===undefined ? null : next)
 *     }
 * }
 */

function doubleIt(head: ListNode | null): ListNode | null {
    head = reverse(head);
    const dummy = new ListNode();
    let cur = dummy;
    let mul = 2;
    let carry = 0;
    while (head) {
        const x = head.val * mul + carry;
        carry = Math.floor(x / 10);
        cur.next = new ListNode(x % 10);
        cur = cur.next;
        head = head.next;
    }
    if (carry) {
        cur.next = new ListNode(carry);
    }
    return reverse(dummy.next);
}

function reverse(head: ListNode | null): ListNode | null {
    const dummy = new ListNode();
    let cur = head;
    while (cur) {
        const next = cur.next;
        cur.next = dummy.next;
        dummy.next = cur;
        cur = next;
    }
    return dummy.next;
}

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