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Reorder List
思路:对列表后半部分逆序排列,然后连接。其中还是需要注意head = cur;cur = cur->next;head->next = tmpNode;相对位置。
/** * Definition for singly-linked list. * struct ListNode { * int val; * ListNode *next; * ListNode(int x) : val(x), next(NULL) {} * }; */ class Solution { public: void reorderList(ListNode* head) { ListNode dummy(-1); dummy.next = head; int count = 0; while(head) { head = head->next; ++count; } head = &dummy; for(int i=0; i<count/2 + count%2; ++i) head = head->next; ListNode *tmp = head->next; head->next = nullptr; //reverse the second half ListNode *cur = nullptr; while(tmp) { ListNode *tmpNode = tmp->next; tmp->next = cur; cur = tmp; tmp = tmpNode; } for(head = dummy.next; head&&cur;) { ListNode *tmpNode = head->next; head->next = cur; head = cur; cur = cur->next; head->next = tmpNode; head = tmpNode; } head = dummy.next; } };
Reorder List
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