四种语言刷算法之反转链表 II

力扣92. 反转链表 II

1、C

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     struct ListNode *next;
 * };
 */
struct ListNode* reverseBetween(struct ListNode* head, int left, int right){
    if(right==left)return head;
    
    struct ListNode* h = (struct ListNode*)malloc(sizeof(struct ListNode));
    h->next = head;
    struct ListNode* q = h;
    for(int i=0;i<left-1;i++){
        q = q->next;
    }
    struct ListNode* p = q;
    q = q->next;
    struct ListNode* r = q->next;
    for(int i=0;i<right-left;i++){
        struct ListNode* t = r->next;
        r->next = q;
        q = r;
        r = t;
    }
    struct ListNode* ll = p->next;
    p->next = q;
    ll->next = r;
    return h->next;

}

2、C++

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode() : val(0), next(nullptr) {}
 *     ListNode(int x) : val(x), next(nullptr) {}
 *     ListNode(int x, ListNode *next) : val(x), next(next) {}
 * };
 */
class Solution {
public:
    ListNode* reverseBetween(ListNode* head, int left, int right) {
        if(right==left)return head;
        ListNode* h = new ListNode(-1);
        h->next = head;
        ListNode * q = h;
        for(int i=0;i<left-1;i++){
            q = q->next;
        }
        ListNode *p = q;
        q = q->next;
        ListNode * r = q->next;
        for(int i=0;i<right-left;i++){
            ListNode* t = r->next;
            r->next = q;
            q = r;
            r = t;
        }
        ListNode* ll = p->next;
        p->next = q;
        ll->next = r;
        return h->next;
    }
};

3、JAVA

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode() {}
 *     ListNode(int val) { this.val = val; }
 *     ListNode(int val, ListNode next) { this.val = val; this.next = next; }
 * }
 */
class Solution {
    public ListNode reverseBetween(ListNode head, int left, int right) {
        if(left==right)return head;
        ListNode h = new ListNode(-1);
        h.next = head;
        ListNode q = h;
        for(int i=0;i<left-1;i++){
            q = q.next;
        }
        ListNode p = q;
        q = q.next;
        ListNode r = q.next;
        for(int i=0;i<right-left;i++){
            ListNode t = r.next;
            r.next = q;
            q = r;
            r = t;
        }
        ListNode ll = p.next;
        p.next = q;
        ll.next = r;
        return h.next;
    }
}

4、Python

# Definition for singly-linked list.
# class ListNode(object):
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
class Solution(object):
    def reverseBetween(self, head, left, right):
        """
        :type head: ListNode
        :type left: int
        :type right: int
        :rtype: ListNode
        """
        if left == right:
            return head
        h = ListNode()
        h.next = head
        q = h
        for i in range(left-1):
            q = q.next
        p = q
        q = q.next
        r = q.next
        for i in range(right-left):
            t = r.next
            r.next = q
            q = r
            r = t
        ll = p.next
        p.next = q
        ll.next = r
        return h.next