#LeetCode:25. Reverse Nodes in k-Group

先想像將 Linked List 試著切成多個 k 長度的 Linked List ,如果有某一個短的 Linked List 不夠 k 長度,則這段不做反轉的動作,其餘則要。

對於每一個 k 長度的 Linked List ,先記住原本的頭和尾以及銜接著這個 Linked List 兩側的節點,接著將中間的部分利用前、中、後三個 pointer 去反轉中間銜接的方式,最後再拿出剛剛記的頭尾銜接完反轉後的順序就大功告成了!

C++(12ms)

/*******************************************************/
/* LeetCode 25. Reverse Nodes in k-Group               */
/* Author: Maplewing [at] knightzone.studio            */
/* Version: 2018/10/19                                 */
/*******************************************************/
/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
  ListNode* reverseKGroup(ListNode* head, int k) {
    if(head == NULL) return head;

    ListNode dummyHead(-1);
    dummyHead.next = head;

    ListNode* previous = &dummyHead;
    ListNode* current = dummyHead.next;

    while(current != NULL){
      ListNode* partialHeadPrevious = previous;
      ListNode* partialTailNext;
      ListNode* partialHead = current;
      ListNode* partialTail;

      ListNode* next = current;
      for(int i = 1 ; i < k ; ++i){
        next = next->next;
        if(next == NULL) return dummyHead.next;
      }

      partialTail = next;
      partialTailNext = next->next;
      for(int i = 0 ; i < k ; ++i){
        next = current->next;
        current->next = previous;
        previous = current;
        current = next;
      }

      partialHeadPrevious->next = partialTail;
      partialHead->next = partialTailNext;
      previous = partialHead;
      current = partialTailNext;
    }

    return dummyHead.next;
  }
};
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