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  • [Leetcode 73] 92 Reverse Linked List II

    Problem:

    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->NULLm = 2 and n = 4,

    return 1->4->3->2->5->NULL.

    Note:
    Given mn satisfy the following condition:
    1 <m < n < length of list.

    Analysis:

    This is an extension problem of reverse a linked list. The first thing to do is to find the starting node of the sub-list to be reversed. Then call the reverse linked-list procedure and adjust the order of nodes. At last, link the new head with the m-1 node together to form the result linked-list. Note in the reverse procedure we connect the sub-list's last node with the n+1 node.

    It's a one-pass and in place algorithm.

    Code:

     1 /**
     2  * Definition for singly-linked list.
     3  * struct ListNode {
     4  *     int val;
     5  *     ListNode *next;
     6  *     ListNode(int x) : val(x), next(NULL) {}
     7  * };
     8  */
     9 class Solution {
    10 public:
    11     ListNode *reverseBetween(ListNode *head, int m, int n) {
    12         // Start typing your C/C++ solution below
    13         // DO NOT write int main() function
    14         if (head == NULL || head->next == NULL)
    15             return head;
    16             
    17         ListNode dmy(-1), *p, *pm;
    18         dmy.next = head;
    19         p = &dmy;
    20         
    21         int cnt = 0;
    22         for (; cnt < m-1; cnt++) {
    23             p = p->next;
    24         }
    25 
    26         // now p points to m-1's node
    27         p->next = reverse(p->next, cnt+1, n);
    28         
    29         return dmy.next;
    30     }
    31     
    32 private:
    33     ListNode *reverse(ListNode *head, int c, int n) {
    34         if (head == NULL || head->next == NULL)
    35             return head;
    36             
    37         ListNode *pre, *cur, *nxt, dmy(-1);
    38         dmy.next = head;
    39         
    40         pre = 0;
    41         cur = dmy.next;
    42         nxt = cur->next;
    43         
    44         while (c <= n) {
    45             nxt = cur->next;
    46             cur->next = pre;
    47             pre = cur;
    48             cur = nxt;
    49             
    50             c++;
    51         }
    52         
    53         head->next = cur;
    54         return pre;
    55     }
    56 };
    View Code
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  • 原文地址:https://www.cnblogs.com/freeneng/p/3205076.html
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