zoukankan      html  css  js  c++  java
  • 【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 ()

    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;
            }
        }
    };

      from here

    Solution 2 ()

    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;
        }
    };
  • 相关阅读:
    初识PL/SQL
    PL/SQL基本语法
    Oracle命令备忘
    工厂模式之二 工厂方法(Factory Method)
    XMLHttpRequest 原始AJAX初步
    DOM元素的innerHTML属性
    如果用JavaScript获取标准下拉框的"选中值"和"选中文本"
    工厂模式之三 抽象工厂(Abstract Factory)模式
    JavaScript中的动态参数
    JavaScript中的闭包初探
  • 原文地址:https://www.cnblogs.com/Atanisi/p/6848336.html
Copyright © 2011-2022 走看看