标签：other push end lintcode reverse http code 题目 dmi
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; } };
标签：other push end lintcode reverse http code 题目 dmi
原文地址：http://www.cnblogs.com/Atanisi/p/6848336.html
