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  • 61. Rotate List

    https://leetcode.com/problems/rotate-list/description/

    Given a linked list, rotate the list to the right by k places, where k is non-negative.

    Example 1:

    Input: 1->2->3->4->5->NULL, k = 2
    Output: 4->5->1->2->3->NULL
    Explanation:
    rotate 1 steps to the right: 5->1->2->3->4->NULL
    rotate 2 steps to the right: 4->5->1->2->3->NULL
    

      

    Example 2:

    Input: 0->1->2->NULL, k = 4
    Output: 2->0->1->NULL
    Explanation:
    rotate 1 steps to the right: 2->0->1->NULL
    rotate 2 steps to the right: 1->2->0->NULL
    rotate 3 steps to the right: 0->1->2->NULL
    rotate 4 steps to the right: 2->0->1->NULL
    

    在看答案之前 自己想了想 大致想法是对的

    只是没有想到用环 这个solution很棒!

    /**
    
    * Definition for singly-linked list.
    * struct ListNode {
    * int val;
    * ListNode *next;
    * ListNode(int x) : val(x), next(NULL) {}
    * };
    */
    class Solution {
    public:
        ListNode* rotateRight(ListNode* head, int k) {
            //k%length(listnode) 就是向前挪几个 
            //fist walk the list and get its length n, save the tail pre_pointer->tail_pre_pointer
            //then compute num = k % n(tail num should be move to the head)
            //index_key = n - num (from n - num to the last n-1 should be move to the head just in order)
            //find the pre_pointer of index_key -> key_pre_pointer
            //tail_pointer->next->next = head_pointer->next,head_pointer = ,key_pre_pointer->next = null key_pre_pointer。。。晕了
            // (以上是自己想法,没有想到用环,下面代码是最佳solution,用到了首尾相连的环 在这种不需要改变前后顺序的题目里面很适合用环 只改变首尾)
    
    
            if(!head) return head;
    
            int len=1; // number of nodes
            ListNode *newH, *tail;
            newH=tail=head;
    
            while(tail->next) // get the number of nodes in the list
            {
            tail = tail->next;
            len++;
            }
            tail->next = head; // circle the link
    
            if(k %= len) 
            {
            for(auto i=0; i<len-k; i++) tail = tail->next; // the tail node is the (len-k)-th node (1st node is head)
            }
            newH = tail->next; 
            tail->next = NULL;
            return newH;
        } 
    };

    https://leetcode.com/problems/rotate-list/discuss/22735/My-clean-C++-code-quite-standard-(find-tail-and-reconnect-the-list)

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  • 原文地址:https://www.cnblogs.com/hozhangel/p/9458958.html
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