[LeetCode] 92. Reverse Linked List II 反向链表II

Reverse a linked list from position m to n. Do it in-place and in one-pass.

For example:
Given 1->2->3->4->5->NULL, m = 2 and n = 4,

return 1->4->3->2->5->NULL.

Note:
Given m, n satisfy the following condition:
1 ≤ m ≤ n ≤ length of list.

206. Reverse Linked List的拓展，这里只是反向一个链表中的一部分。先建立一个新list node: dummy，用dummy链接m之前不用反向的node，然后反向m~n的节点，最后在把反向的连接起来。

Java:

/**
 * Definition for ListNode
 * public class ListNode {
 *     int val;
 *     ListNode next;
 * }
 */
public class Solution {
    public ListNode reverseBetween(ListNode head, int m, int n) {
        if (m >= n || head == null) {
            return head;
        }
        
        ListNode dummy = new ListNode(0);
        dummy.next = head;
        head = dummy;
        
        for (int i = 1; i < m; i++) {
            if (head == null) {
                return null;
            }
            head = head.next;
        }
        
        ListNode premNode = head;
        ListNode mNode = head.next;
        ListNode nNode = mNode, postnNode = mNode.next;
        for (int i = m; i < n; i++) {
            if (postnNode == null) {
                return null;
            }
            ListNode temp = postnNode.next;
            postnNode.next = nNode;
            nNode = postnNode;
            postnNode = temp;
        }
        mNode.next = postnNode;
        premNode.next = nNode;
        
        return dummy.next;
    }
}

Python:

class Solution:
    def reverse(self, head):
        prev = None
        while head != None:
            next = head.next
            head.next = prev
            prev = head
            head = next
        return prev

    def findkth(self, head, k):
        for i in xrange(k):
            if head is None:
                return None
            head = head.next
        return head

    def reverseBetween(self, head, m, n):
        dummy = ListNode(-1, head)
        mth_prev = self.findkth(dummy, m - 1)
        mth = mth_prev.next
        nth = self.findkth(dummy, n)
        nth_next = nth.next
        nth.next = None

        self.reverse(mth)
        mth_prev.next = nth
        mth.next = nth_next
        return dummy.next

Python:

class Solution:
    def reverseBetween(self, head, m, n):
        dummyNode = ListNode(0)
        p = dummyNode
        q = head

        for x in range(m - 1):
            p.next = q
            q = q.next
            p = p.next

        start = None
        end = q
        next = None
        for x in range(m, n + 1):
            next = q.next
            q.next = start
            start = q
            q = next

        p.next = start
        end.next = next
        return dummyNode.next

C++:

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode* reverseBetween(ListNode* head, int m, int n) {
        ListNode dummy{0};
        dummy.next = head;

        auto *prev = &dummy;

        for (int i = 0; i < m - 1; ++i) {
            prev = prev->next;
        }

        auto *head2 = prev;

        prev = prev->next;
        auto *cur = prev->next;

        for (int i = m; i < n; ++i) {
            prev->next = cur->next;  // Remove cur from the list.
            cur->next = head2->next; // Add cur to the head.
            head2->next = cur;       // Add cur to the head.
            cur = prev->next;        // Get next cur.
        }

        return dummy.next;
    }
};

　　

　　

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[LeetCode] 206. Reverse Linked List 反向链表

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