leetcode - Reorder List
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}
.
个人思路:
1,找到链表的中点,断开分为两部分,对链表的后半部分进行反转,然后再合并这两部分链表即可
代码:
1 #include <stddef.h> 2 3 struct ListNode 4 { 5 int val; 6 ListNode *next; 7 ListNode(int x) : val(x), next(NULL) {} 8 }; 9 10 class Solution { 11 public: 12 void reorderList(ListNode *head) { 13 if (!head || !head->next) 14 { 15 return; 16 } 17 18 int length = 0; 19 ListNode *current = head; 20 21 while (current) 22 { 23 current = current->next; 24 ++length; 25 } 26 27 ListNode *firstHead = head; 28 int firstLength = length / 2 + length % 2; 29 ListNode *secondHead = NULL; 30 current = firstHead; 31 32 for (int i = 0; i < firstLength; ++i) 33 { 34 current = current->next; 35 } 36 secondHead = current; 37 38 //对second链表进行反转 39 ListNode dummy(-1); 40 dummy.next = secondHead; 41 ListNode *pre = secondHead; 42 current = secondHead->next; 43 44 while (current) 45 { 46 pre->next = current->next; 47 current->next = dummy.next; 48 dummy.next = current; 49 current = pre->next; 50 } 51 52 //合并first链表和second链表 53 ListNode *firstCur = firstHead; 54 ListNode *secondCur = dummy.next; 55 ListNode *temp = NULL; 56 57 while (firstCur && secondCur) 58 { 59 temp = secondCur->next; 60 secondCur->next = firstCur->next; 61 firstCur->next = secondCur; 62 63 firstCur = secondCur->next; 64 secondCur = temp; 65 } 66 67 if (firstCur) 68 { 69 firstCur->next = NULL; 70 } 71 } 72 };
posted on 2014-09-06 11:32 laihaiteng 阅读(168) 评论(0) 编辑 收藏 举报