【Lintcode】099.Reorder List

题目:

Given a singly linked list L: L0 → L1 → … → Ln-1 → Ln

reorder it to: L0 → Ln → L1 → Ln-1 → L2 → Ln-2 → …

Example

Given 1->2->3->4->null, reorder it to 1->4->2->3->null.

题解:

   Spliting the list from the middle into two lists. One from head to middle, and the other from middle to the end. Then we reverse the second list. Finally we merge these two lists

Solution 1 ()

复制代码
class Solution {
public:
    void reorderList(ListNode *head) {
        if (!head || !head->next) {
            return;
        }
        ListNode* mid = findMiddle(head);
        ListNode* right = reverse(mid->next);
        mid->next = nullptr;
        merge(head, right);
    }
    ListNode* findMiddle(ListNode* head) {
        ListNode* slow = head;
        ListNode* fast = head->next;
        while (fast && fast->next) {
            slow = slow->next;
            fast = fast->next->next;
        }
        
        return slow;
    }
    ListNode* reverse(ListNode* head) {
        if (!head || !head->next) {
            return head;
        }
        
        ListNode* pre = nullptr;
        while (head) {
            ListNode* tmp = head->next;
            head->next = pre;
            pre = head;
            head = tmp;
        }
        return pre;
    }
    void merge(ListNode* left, ListNode* right) {
        ListNode* dummy = new ListNode(-1);
        int idx = 0;
        while (left && right) {
            if (idx % 2 == 0) {
                dummy->next = left;
                left = left->next;
            } else {
                dummy->next = right;
                right = right->next; 
            }
            dummy = dummy->next;
            ++idx;
        }    
        if (left) {
            dummy->next = left;
        } else {
            dummy->next = right;
        }
    }
};
复制代码

  from here

Solution 2 ()

复制代码
class Solution {
public:
    /**
     * @param head: The first node of linked list.
     * @return: void
     */
    void reorderList(ListNode *head) {
        // write your code here
        if (head == NULL)
            return;
        
        vector<ListNode*> nodes;
        ListNode* iter = head;
        while(iter != NULL)
        {
            nodes.push_back(iter);
            iter = iter->next;
        }
        
        int LEN = nodes.size();
        int left = 0;
        int right = LEN -1;
        while(left < right)
        {
            nodes[left]->next = nodes[right];
            nodes[right--]->next = nodes[++left];
        }
        nodes[left]->next = NULL;
    }
};
复制代码

 

posted @   Vincent丶丶  阅读(212)  评论(0编辑  收藏  举报
努力加载评论中...
点击右上角即可分享
微信分享提示