三步翻转法

Posted 2022-04-13 roni-i

class Solution 
    public void reverse(char[] ch, int l ,int r) 
        for ( ; l<r; l++,r--)
            char t = ch[l];
            ch[l] = ch[r];
            ch[r] = t;
        
    
    public void fun(String str, int n, int m)
        if(str == null || str.length()==0) return;
        m %= n;
        char[] ch = str.toCharArray();
        reverse(ch, 0, m-1);
        reverse(ch, m, n-1);
        reverse(ch, 0, n-1);
        for(int i=0; i<ch.length; i++)
            System.out.print(ch[i]);
        
    
    public static void main(String[] args) 
        String s = "abcdef";
        new Solution().fun(s,6,4); //前四个字符移到末尾：efabcd

    

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1. Reverse Nodes in k-Group
   Hard

1244

257

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Given a linked list, reverse the nodes of a linked list k at a time and return its modified list.

k is a positive integer and is less than or equal to the length of the linked list. If the number of nodes is not a multiple of k then left-out nodes in the end should remain as it is.

Example:

Given this linked list: 1->2->3->4->5

For k = 2, you should return: 2->1->4->3->5

For k = 3, you should return: 3->2->1->4->5

Note:

Only constant extra memory is allowed.
You may not alter the values in the list‘s nodes, only nodes itself may be changed.

/**
 * Definition for singly-linked list.
 * public class ListNode 
 *     int val;
 *     ListNode next;
 *     ListNode(int x)  val = x; 
 * 
 */
class Solution 
    public ListNode reverseKGroup(ListNode head, int k) 
        ListNode dummy = new ListNode(0);
        dummy.next = head;
        ListNode pre = dummy;
        ListNode end = dummy;
        while(end.next != null)
            for(int i=0; i<k && end != null; i++) end = end.next;
            if(end == null) break;
            ListNode start = pre.next;
            ListNode next = end.next;
            end.next = null;
            pre.next = reverse(start);
            start.next = next;
            pre = start;
            end = pre;
        
        return dummy.next;
    
    public ListNode reverse(ListNode head)
        ListNode pre = null;
        while(head != null)
            ListNode next = head.next;
            head.next = pre;
            pre = head;
            head = next;
        
        return pre;