1 /** 2 * Definition for singly-linked list. 3 * struct ListNode { 4 * int val; 5 * ListNode *next; 6 * ListNode(int x) : val(x), next(NULL) {} 7 * }; 8 */ 9 class Solution { 10 public: 11 12 ListNode* reverse(ListNode* head) { 13 if (head == NULL) { 14 return NULL; 15 } 16 ListNode* pre = NULL; 17 ListNode* cur = head; 18 while (cur != NULL) { 19 ListNode* tmp = cur->next; 20 cur->next = pre; 21 pre = cur; 22 cur = tmp; 23 } 24 return pre; 25 } 26 27 28 void reorderList(ListNode *head) { 29 if (head == NULL || head->next == NULL) { 30 return; 31 } 32 int num = 0; 33 ListNode* cur = head; 34 while (cur != NULL) { 35 num++; 36 cur = cur->next; 37 } 38 // seperate list, find middle node as the head of the new list 39 int ridx = num / 2; 40 ListNode* rhead = NULL; 41 cur = head; 42 for (int i=0; true; i++) { 43 if (i == ridx - 1) { 44 rhead = cur->next; 45 cur->next = NULL; 46 break; 47 } 48 cur = cur->next; 49 } 50 51 rhead = reverse(rhead); 52 53 cur = head; 54 head = head->next; 55 56 cur->next = rhead; 57 cur = rhead; 58 rhead = rhead->next; 59 60 while (head != NULL && rhead != NULL) { 61 cur->next = head; 62 cur = head; 63 head = head->next; 64 65 cur->next = rhead; 66 cur = rhead; 67 rhead = rhead->next; 68 } 69 cur->next = rhead; // there must be only one node left(odd case) or NULL(even case) 70 } 71 72 };
这题跟Copy List with Random Pointer 那题类似在生成结果前都对原先的链表结构进行了调整，这里就是把链表的后半段给逆序了一下，这样我们就能轻松的取出节点对(0, n), (1, n-1)...了（序号是原先未修改时的，[0, n]）
第二轮：
Given a singly linked list L: L0→L1→…→Ln-1→Ln,
reorder it to: L0→Ln→L1→Ln-1→L2→Ln-2→…
You must do this in-place without altering the nodes' values.
For example,
Given {1,2,3,4}, reorder it to {1,4,2,3}.
1 /** 2 * Definition for singly-linked list. 3 * struct ListNode { 4 * int val; 5 * ListNode *next; 6 * ListNode(int x) : val(x), next(NULL) {} 7 * }; 8 */ 9 // 10:26 10 class Solution { 11 public: 12 void reorderList(ListNode *head) { 13 if (head == NULL) { 14 return; 15 } 16 ListNode fakeHead(0); 17 fakeHead.next = head; 18 ListNode* slow = &fakeHead; 19 ListNode* fast = &fakeHead; 20 while (fast != NULL && fast->next != NULL) { 21 slow = slow->next; 22 fast = fast->next->next; 23 } 24 ListNode* second = slow->next; 25 slow->next = NULL; 26 27 ListNode* pre = NULL; 28 ListNode* cur = second; 29 while (cur != NULL) { 30 ListNode* tmp = cur->next; 31 cur->next = pre; 32 pre = cur; 33 cur = tmp; 34 } 35 ListNode* ahead = fakeHead.next; 36 ListNode* bhead = pre; 37 38 fakeHead.next = NULL; 39 ListNode* last = &fakeHead; 40 41 while (ahead != NULL && bhead != NULL) { 42 last->next = ahead; 43 last = ahead; 44 ahead = ahead->next; 45 last->next = bhead; 46 last = bhead; 47 bhead = bhead->next; 48 } 49 last->next = ahead; 50 } 51 };
取中间点的方法改进了一下