C++ 程序减去表示为链接列表的两个数字

给定两个表示两个大正数的链表。从较大的数字中减去较小的数字，并将差值作为链表返回。请注意，输入列表可以是任何顺序，但我们总是需要从较大的列表中减去较小的列表。
可以假设输入列表中没有额外的前导零。
例子：

Input: l1 = 1 -> 0 -> 0 -> NULL,  l2 = 1 -> NULL
Output: 0->9->9->NULL
Explanation: Number represented as 
lists are 100 and 1, so 100 - 1 is 099

Input: l1 = 7-> 8 -> 6 -> NULL,  l2 = 7 -> 8 -> 9 NULL
Output: 3->NULL
Explanation: Number represented as 
lists are 786 and  789, so 789 - 786 is 3, 
as the smaller value is subtracted from 
the larger one.

方法：以下是步骤。

计算给定两个链表的大小。
如果大小不同，则在较小的链表中附加零。
如果大小相同，请按照以下步骤操作：
1. 找到较小值的链表。
2. 从较大的链表中一一减去较小链表的节点。减法时跟踪借位。

以下是上述方法的实现。

C++

// C++ program to subtract smaller valued 
// list from larger valued list and return 
// result as a list.
#include 
using namespace std;
  
// A linked List Node
struct Node 
{
    int data;
    struct Node* next;
};
  
// A utility which creates 
// Node.
Node* newNode(int data)
{
    Node* temp = new Node;
    temp->data = data;
    temp->next = NULL;
    return temp;
}
  
/* A utility function to get length
   of linked list */
int getLength(Node* Node)
{
    int size = 0;
    while (Node != NULL) 
    {
        Node = Node->next;
        size++;
    }
    return size;
}
  
/* A Utility that padds zeros in 
   front of the Node, with the 
   given diff */
Node* paddZeros(Node* sNode, 
                int diff)
{
    if (sNode == NULL)
        return NULL;
  
    Node* zHead = newNode(0);
    diff--;
    Node* temp = zHead;
    while (diff--) 
    {
        temp->next = newNode(0);
        temp = temp->next;
    }
    temp->next = sNode;
    return zHead;
}
  
/* Subtract LinkedList Helper is a 
   recursive function, move till the 
   last Node,  and subtract the digits 
   and create the Node and return the 
   Node. If d1 < d2, we borrow the number 
   from previous digit. */
Node* subtractLinkedListHelper(Node* l1, 
                               Node* l2, 
                               bool& borrow)
{
    if (l1 == NULL && 
        l2 == NULL && borrow == 0)
        return NULL;
  
    Node* previous = subtractLinkedListHelper(
                     l1 ? l1->next : NULL,
                     l2 ? l2->next : NULL, borrow);
  
    int d1 = l1->data;
    int d2 = l2->data;
    int sub = 0;
  
    /* If you have given the value value 
       to next digit then reduce the d1 by 1 */
    if (borrow) 
    {
        d1--;
        borrow = false;
    }
  
    /* If d1 < d2, then borrow the number 
       from previous digit. Add 10 to d1 
       and set borrow = true; */
    if (d1 < d2) 
    {
        borrow = true;
        d1 = d1 + 10;
    }
  
    // Subtract the digits 
    sub = d1 - d2;
  
    // Create a Node with sub value 
    Node* current = newNode(sub);
  
    // Set the Next pointer as Previous 
    current->next = previous;
  
    return current;
}
  
/* This API subtracts two linked lists 
   and returns the linked list which 
   shall  have the subtracted result. */
Node* subtractLinkedList(Node* l1, 
                         Node* l2)
{
    // Base Case.
    if (l1 == NULL && l2 == NULL)
        return NULL;
  
    // In either of the case, get the 
    // lengths of both Linked list.
    int len1 = getLength(l1);
    int len2 = getLength(l2);
  
    Node *lNode = NULL, *sNode = NULL;
  
    Node* temp1 = l1;
    Node* temp2 = l2;
  
    // If lengths differ, calculate the 
    // smaller Node and padd zeros for 
    // smaller Node and ensure both larger 
    // Node and smaller Node has equal length.
    if (len1 != len2) 
    {
        lNode = len1 > len2 ? l1 : l2;
        sNode = len1 > len2 ? l2 : l1;
        sNode = paddZeros(sNode, 
                          abs(len1 - len2));
    }
  
    else
    {
        // If both list lengths are equal, 
        // then calculate the larger and 
        // smaller list. If 5-6-7 & 5-6-8 
        // are linked list, then walk through 
        // linked list at last Node as 7 < 8, 
        // larger Node is 5-6-8 and smaller 
        // Node is 5-6-7.
        while (l1 && l2) 
        {
            if (l1->data != l2->data) 
            {
                lNode = (l1->data > l2->data ? 
                         temp1 : temp2);
                sNode = (l1->data > l2->data ? 
                         temp2 : temp1);
                break;
            }
            l1 = l1->next;
            l2 = l2->next;
        }
    }
      // If both lNode and sNode still 
      // have NULL value, then this means 
      // that the  value of both of the 
      // given linked lists is the same 
      // and hence we can directly return 
      // a node with value 0.
    if(lNode==NULL&&sNode==NULL)
    {
      return newNode(0);
    }
  
    // After calculating larger and smaller 
    // Node, call subtractLinkedListHelper 
    // which returns the subtracted linked list.
    bool borrow = false;
    return subtractLinkedListHelper(lNode, 
                                    sNode, borrow);
}
  
/* A utility function to print 
   linked list */
void printList(struct Node* Node)
{
    while (Node != NULL) 
    {
        printf("%d ", Node->data);
        Node = Node->next;
    }
    printf("");
}
  
// Driver code
int main()
{
    Node* head1 = newNode(1);
    head1->next = newNode(0);
    head1->next->next = newNode(0);
    Node* head2 = newNode(1);
    Node* result = subtractLinkedList(head1, 
                                      head2);
    printList(result);
    return 0;
}

输出：

0 9 9

复杂性分析：

时间复杂度： O(n)。
因为不需要嵌套遍历链表。
辅助空间： O(n)。
如果考虑递归堆栈空间，则需要 O(n) 空间。

有关更多详细信息，请参阅有关以链接列表表示的减去两个数字的完整文章！