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📜  Java程序将链表的子列表从位置M到N向右旋转K位

📅  最后修改于: 2022-05-13 01:55:15.233000             🧑  作者: Mango

Java程序将链表的子列表从位置M到N向右旋转K位

给定一个链表和两个位置“m”和“n”。任务是将子列表从位置 m 旋转到 n，向右旋转 k 个位置。

例子：

方法：为了旋转从 m 到 n 元素的给定子列表，将列表从第 (n-k+1)个节点移动到第 n个节点到子列表的开头以完成旋转。
如果 k 大于子列表的大小，那么我们将取其与子列表大小的模数。因此，使用指针和计数器遍历列表，我们将保存第 (m-1)个节点，然后使其指向第 (n-k+1)个节点，从而将第 (n-k+1)个节点带到子列表的开始（前）。
同样，我们将保存第 m个节点，然后让第 n个节点指向它。为了保持列表的其余部分完好无损，我们将使 (nk) th节点指向 n 的下一个节点（可能为 NULL）。最后我们将得到 k 次右旋子列表。

下面是上述方法的实现：

Java
// Java implementation of the above approach 
import java.util.*;
  
class Solution
{
  
    
// Definition of node of linkedlist 
static class ListNode { 
    int data; 
     ListNode next; 
}
    
// This function take head pointer of list, start and 
// end points of sublist that is to be rotated and the 
// number k and rotate the sublist to right by k places. 
static void rotateSubList(ListNode A, int m, int n, int k) 
{ 
    int size = n - m + 1; 
    
    // If k is greater than size of sublist then  
    // we will take its modulo with size of sublist 
    if (k > size) { 
        k = k % size; 
    } 
    
    // If k is zero or k is equal to size or k is 
    // a multiple of size of sublist then list  
    // remains intact 
    if (k == 0 || k == size) { 
        ListNode head = A; 
        while (head != null) { 
            System.out.print( head.data); 
            head = head.next; 
        } 
        return; 
    } 
    
    ListNode link = null;  // m-th node 
    if (m == 1) { 
        link = A; 
    } 
    
    // This loop will traverse all node till 
    // end node of sublist.     
    ListNode c = A;  // Current traversed node 
    int count = 0;  // Count of traversed nodes 
    ListNode end = null;   
    ListNode pre = null; // Previous of m-th node 
    while (c != null) { 
        count++; 
    
        // We will save (m-1)th node and later 
        // make it point to (n-k+1)th node 
        if (count == m - 1) { 
            pre = c; 
            link = c.next; 
        } 
        if (count == n - k) { 
            if (m == 1) { 
                end = c; 
                A = c.next; 
            } 
            else { 
                end = c; 
    
                // That is how we bring (n-k+1)th 
                // node to front of sublist. 
                pre.next = c.next; 
            } 
        } 
    
        // This keeps rest part of list intact. 
        if (count == n) { 
            ListNode d = c.next; 
            c.next = link; 
            end.next = d; 
            ListNode head = A; 
            while (head != null) { 
                System.out.print( head.data+" "); 
                head = head.next; 
            } 
            return; 
        } 
        c = c.next; 
    } 
} 
    
// Function for creating and linking new nodes 
static ListNode push( ListNode head, int val) 
{ 
     ListNode new_node = new ListNode(); 
    new_node.data = val; 
    new_node.next = (head); 
    (head) = new_node; 
    return head;
} 
    
// Driver code 
public static void main(String args[])
{ 
     ListNode head = null; 
    head =push(head, 70); 
    head =push(head, 60); 
    head =push(head, 50); 
    head =push(head, 40); 
    head =push(head, 30); 
    head =push(head, 20); 
    head =push(head, 10); 
    ListNode tmp = head; 
    System.out.print("Given List: "); 
    while (tmp != null) { 
        System.out.print( tmp.data + " "); 
        tmp = tmp.next; 
    } 
    System.out.println();
    
    int m = 3, n = 6, k = 2; 
    System.out.print( "After rotation of sublist: "); 
    rotateSubList(head, m, n, k); 
    
} 
}
  
// This code is contributed
// by Arnab Kundu

输出： 
Given List: 10 20 30 40 50 60 70 
After rotation of sublist: 10 20 50 60 30 40 70

详情请参阅完整文章将链表的子列表从位置 M 向右旋转 K 位！