Paste_Image.png

My code:

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
public class Solution {
    public ListNode rotateRight(ListNode head, int k) {
        if (head == null || k < 0)
            return head;
        else if (head.next == null)
            return head;
        
        ListNode dummy = new ListNode(Integer.MIN_VALUE);
        dummy.next = head;
        ListNode tail = head;
        int total = 1;
        while (tail.next != null) {
            tail = tail.next;
            total++;
        }
        k = k % total;
        if (k == 0)
            return dummy.next;
        
        ListNode curr = dummy;
        for (int i = 0; i < (total - k); i++)
            curr = curr.next;
        if (curr.next == null)
            return dummy.next;
        ListNode newHead = curr.next;
        curr.next = null;
        tail.next = dummy.next;
        dummy.next = newHead;
        return dummy.next;
    }
    
    public static void main(String[] args) {
        Solution test = new Solution();
        ListNode n1 = new ListNode(1);
        ListNode n2 = new ListNode(2);
        ListNode n3 = new ListNode(3);
        ListNode n4 = new ListNode(4);
        ListNode n5 = new ListNode(5);
        n1.next = n2;
        n2.next = n3;
        n3.next = n4;
        n4.next = n5;
        System.out.println(test.rotateRight(n1, 3));
    }
}

My test result:

Paste_Image.png

這次題目也不難。先遍歷一遍找出總個(gè)數(shù) total
然后根據(jù) total - k 找到那個(gè)斷裂點(diǎn)。斷開(kāi)。在用dummy和新的頭接上。
原來(lái)的頭接到tail后面去，就差不多了。
主要我是沒(méi)能理解， k >= total 是什么意義。
查了之后發(fā)現(xiàn)，做一個(gè) k = k % total 處理就行了。

**
總結(jié)： LinkedList rotate
**

Anyway, Good luck, Richardo!

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
public class Solution {
    public ListNode rotateRight(ListNode head, int k) {
        if (head == null)
            return null;
        int n = 1;
        ListNode temp = head;
        ListNode tail = head;
        while (temp.next != null) {
            n++;
            temp = temp.next;
            tail = temp;
        }
        k = k % n;
        if (k == 0)
            return head;
        temp = head;
        for (int i = 1; i < n - k; i++) {
            temp = temp.next;
        }
        ListNode newHead = temp.next;
        temp.next = null;
        tail.next = head;
        return newHead;
    }
    
    public static void main(String[] args) {
        ListNode n1 = new ListNode(1);
        ListNode n2 = new ListNode(2);
        ListNode n3 = new ListNode(3);
        ListNode n4 = new ListNode(4);
        ListNode n5 = new ListNode(5);
        n1.next = n2;
        n2.next = n3;
        n3.next = n4;
        n4.next = n5;
        Solution test = new Solution();
        ListNode head = test.rotateRight(n1, 2);
        ListNode temp = head;
        while (temp != null) {
            System.out.print(temp.val + "->");
            temp = temp.next;
        }
    }
}

有一個(gè) corner case沒(méi)有考慮到，當(dāng)k= 0 時(shí)，這個(gè)時(shí)候是不需要旋轉(zhuǎn)的，如果強(qiáng)制旋轉(zhuǎn)， newHead = tail.next -> null. 會(huì)出錯(cuò)。

Anyway, Good luck, Richardo!

My code:

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
public class Solution {
    public ListNode rotateRight(ListNode head, int k) {
        if (k < 0 || head == null) {
            return null;
        }
        
        ListNode tail = head;
        int total = 0;
        while (tail.next != null) {
            total++;
            tail = tail.next;
        }
        total++;
        k = k % total;
        if (k == 0) {
            return head;
        }
        int counter = total - k - 1;
        ListNode curr = head;
        while (counter > 0) {
            curr = curr.next;
            counter--;
        }
        
        ListNode newHead = curr.next;
        curr.next = null;
        tail.next = head;
        return newHead;
    }
}

k = 0 的時(shí)候，不需要rotate，直接返回就行。
Anyway, Good luck, Richardo! -- 08/16/2016