39：Reorder List

最新推荐文章于 2019-12-23 17:34:46 发布

weishenmetlc

最新推荐文章于 2019-12-23 17:34:46 发布

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分类专栏： leetcode

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本文介绍了一种链表重组算法，该算法将链表分为两部分并逆序后半部分，再交替合并为新链表。提供了完整的C++实现代码。

题目：Given a singly linked list L : $L_{0} -> L_{1} -> ... -> L_{n - 1} -> L_{n}$ , reorder it to: $L_{0} -> L_{n} -> L_{1} -> L_{n -1} -> L_{2} -> L_{n - 2} ->$ …
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}.

解析：将链表从中间断开，然后将后半段链表逆序，最后将前半段链表以及逆过序后的后半段链表合并在一起

解题代码如下：

class Solution {
public:
        void reorderList(ListNode *head) {
                if (!head || !head -> next) return;
                //计算链表总长度
                int len = 0;
                ListNode *cur = head;
                while (cur != nullptr) {
                        ++len;
                        cur = cur -> next;
                }

                int mid = len % 2 != 0 ? len / 2 + 1 : len / 2;
                ListNode* mid_ptr = head;
                for (int i = 0; i != mid - 1; ++i)
                        mid_ptr = mid_ptr -> next;

                // 从中间将原先链表分成两个链表
                ListNode* head2 = mid_ptr -> next;
                mid_ptr -> next = nullptr;
                // 将后半段链表逆序
                head2 = reverseList(head2);

                //将这两个链表再次合并成一个链表
                ListNode *cur1 = head, *cur2 = head2;
                while (cur2 != nullptr) {
                        ListNode* tmp = cur2 -> next;
                        cur2 -> next = cur1 -> next;
                        cur1 -> next = cur2;
                        cur1  = cur1 -> next -> next;
                        cur2 = tmp;
                }
        }
private:
        ListNode* reverseList(ListNode* head) {
                if (!head) return head;
                ListNode dummy{-1};
                dummy.next = head;

                ListNode* cur = head;
                while (cur -> next != nullptr) {
                        ListNode* tmp = cur -> next;
                        cur -> next = tmp -> next;
                        tmp -> next = dummy.next;
                        dummy.next = tmp;
                }
                return head;
        }
};