61 Rotate List

最新推荐文章于 2020-10-07 20:15:39 发布

weixin_39145266

最新推荐文章于 2020-10-07 20:15:39 发布

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分类专栏： LeetCode # Medium # Solved # LinkedList # TwoPointers

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本文提供了一种解决方案，用于将链表向右旋转k个位置。通过两段不同的代码实现了这一功能，一种方法是寻找链表的末尾并计算其长度，然后确定实际需要旋转的位置；另一种方法则将链表连接成环形，再进行旋转。

1 题目

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

2 尝试解

class Solution {
public:
    ListNode* rotateRight(ListNode* head, int k) {
        if(head == NULL) return head;
        ListNode* slow = head;
        ListNode* fast = head;
        int size = 1;
        while(fast->next !=NULL){  //fast to the end, get the size of list
            fast = fast->next;
            size ++;
        }
        if(size <= k) {  
            k = k % (size);
        }
        int count = 1;
        while(count < size - k){ //slow to the breakpoint
            slow = slow->next;
            count++;
        }
        
        if(slow != fast){
            ListNode* start = slow->next;
            fast->next = head;
            slow->next = NULL;
            return start;
        }
        return head;
    }
};

3 标准解

class Solution {
public:
    ListNode* rotateRight(ListNode* head, int k) {
        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;
    }
};