通过用K的最接近倍数替换每个节点来修改链接列表

给定一个由N个节点和一个整数K组成的单链表L ，任务是将给定链表的每个节点的值修改为K的最接近倍数，但不超过该节点的值。

例子：

Input: LL: 1 -> 2 -> 3 -> 5, K = 2 
Output: 0 -> 2 -> 2 -> 4
Explanation:
The resultant LL has values having less than the original LL value and is a multiple of K(= 2).

Input: LL:13 -> 14 -> 26 -> 29 -> 40, K = 13
Output: 13 -> 13 -> 26 -> 26 -> 39

方法：可以通过遍历给定的链表来解决给定的问题，并且对于所遍历的每个节点，找到节点值除以K的底值，例如val ，并将节点值更新为val * K。

下面是上述方法的实现：

C++

// C++ program for the above approach
  
#include 
using namespace std;
  
// Structure of node
struct node {
    int data;
    node* next;
};
  
// Function to replace the node N
// by the nearest multiple of K 
node* EvalNearestMult(node* N, int K)
{
    node* temp = N;
    int t;
  
    // Traverse the Linked List
    while (N != NULL) {
  
        // If data is less than K
        if (N->data < K)
            N->data = 0;
  
        else {
  
            // If the data of current
            // node is same as K
            if (N->data == K)
                N->data = K;
  
            // Otherwise change the value
            else {
                N->data = (N->data / K) * K;
            }
        }
  
        // Move to the next node
        N = N->next;
    }
  
    // Return the updated LL
    return temp;
}
  
// Function to print the nodes of
// the Linked List
void printList(node* N)
{
    // Traverse the LL
    while (N != NULL) {
  
        // Print the node's data
        cout << N->data << "  ";
        N = N->next;
    }
}
  
// Driver Code
int main()
{
    // Given Linked List
    node* head = NULL;
    node* second = NULL;
    node* third = NULL;
  
    head = new node();
    second = new node();
    third = new node();
  
    head->data = 3;
    head->next = second;
    second->data = 4;
    second->next = third;
    third->data = 8;
    third->next = NULL;
  
    node* t = EvalNearestMult(head, 3);
  
    printList(t);
  
    return 0;
}

输出：

3  3  6

时间复杂度： O(N)
辅助空间： O(1)