Reorder List
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原题如下:
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}
.
解题思路:
(1)首先判断链表是否为空,是否只含有一个结点等。
(2)找到链表的中间结点,这个可以通过“快慢”指针的方法找到。然后就地逆置后半部分链表。
(3) 把前半部分的链表和逆置后的链表“逐一”串起来,即可AC此题。
C++代码如下:
/** * Definition for singly-linked list. * struct ListNode { * int val; * ListNode *next; * ListNode(int x) : val(x), next(NULL) {} * }; */class Solution {public: ListNode* reverseLinkList(ListNode* head){ if(!head && !head->next) return head; ListNode* first, *second, *temp; first = head; second = head->next; while(second){ temp = second->next; second->next = first; first = second; second = temp; } head->next = NULL; return first; } void reorderList(ListNode* head) { if(!head || !head->next){ return; } ListNode* slow, *fast; slow = fast = head; while(fast && fast->next){ slow = slow->next; fast = fast->next->next; } ListNode* tail = reverseLinkList(slow); ListNode* tempHead = head, *temp1, *temp2; while(tempHead && tempHead->next && tail && tail->next){ temp1 = tempHead->next; tempHead->next = tail; temp2 = tail->next; tail->next = temp1; tempHead = temp1; tail = temp2; } }};
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