92. Reverse Linked List II

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 Author:hozhangel

Time:2018-08-12 11:45:12

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92. Reverse Linked List II

similar problem: 206. Reverse Linked List

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

Note: 1 ≤ m ≤ n ≤ length of list.

Example:

Input: 1->2->3->4->5->NULL, m = 2, n = 4
Output: 1->4->3->2->5->NULL

just like the 206-th ,and add the count

//自己写的代码 在206th 基础上改动 ^^：

/**
 * 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) {
        if (head == nullptr || head->next == nullptr) {
            return head;
        }
        ListNode* pre = new ListNode(0);
        pre->next = head;
        ListNode* pre_new = pre;
        ListNode* p = pre_new->next;
        ListNode* q = p->next;
        int mm = 2;
        while (q&&mm<=n) {
            
            if(mm<=m){       //find the begin node to reverse
                pre_new=pre_new->next;
                p = pre_new->next;
                q = p->next;
            }
            else{            //just reverse
                ListNode* pt = q;
                p->next = pt->next;
                pt->next = pre_new->next;
                pre_new->next = pt;
                q=p->next;
            }
            mm++;
            
        }

        return pre->next;
    }
};

 

其他值得学习的best-solution: 递归C++-simple-solution-easy-to-understand

class Solution {
public:

ListNode* reverseBetween(ListNode* head, int m, int n) {
    ListNode* curNode = head;
    for(int i=0;i<m-1;++i) curNode = curNode->next;
    reverseBetween(curNode,n-m);
    return head;
}

void reverseBetween(ListNode* head, int length){
    if(length<=0) return;
   
    ListNode* lastNode = head;
    for(int i=0;i!=length;++i) lastNode = lastNode->next;
    swap(head->val,lastNode->val);
    reverseBetween(head->next,length-2);
}
};