标签：des style blog color 使用 io for ar
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}.
思路：遍历输入链表，将每个元素存入数组中，然后分别向后、向前遍历数组，对链表进行交叉合并。
class Solution { public: void reorderList( ListNode *head ) { if( !head || !head->next ) { return; } vector<ListNode*> nodes; ListNode *curr = head; while( curr ) { nodes.push_back( curr ); curr = curr->next; } int begin = 0, end = (int)nodes.size()-1; while( begin+1 < end ) { nodes[end]->next = nodes[begin]->next; nodes[begin]->next = nodes[end]; ++begin; --end; } nodes[end]->next = 0; return; } };
上述方法使用了与链表元素总数等长的数组。为了满足常空间复杂度要求，参考网上的思路：从中间部分将输入链表分割为前后两个链表，然后将后一个链表翻转，最后交叉合并两个链表。
1 class Solution { 2 public: 3 void reorderList( ListNode *head ) { 4 if( !head || !head->next ) { return; } 5 ListNode *slow = head, *fast = head->next; 6 while( fast->next ) { 7 slow = slow->next; 8 fast = fast->next; 9 if( fast->next ) { fast = fast->next; } 10 } 11 fast = slow->next; 12 slow->next = 0; 13 fast = reverse( fast ); 14 slow = head; 15 while( slow && fast ) { 16 ListNode *tmp = slow->next; 17 slow->next = fast; 18 fast = fast->next; 19 slow->next->next = tmp; 20 slow = tmp; 21 } 22 return; 23 } 24 private: 25 ListNode* reverse( ListNode *head ) { 26 ListNode *next = head->next; 27 head->next = 0; 28 while( next ) { 29 ListNode *tmp = next->next; 30 next->next = head; 31 head = next; 32 next = tmp; 33 } 34 return head; 35 } 36 };
标签：des style blog color 使用 io for ar
原文地址：http://www.cnblogs.com/moderate-fish/p/3904295.html
