Leetcode NO.143 Reorder List

本题题目要求如下：

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}

.
之前做这道题感觉还是挺难的，但是最近做了几道reverse linked list相关的题，就不难了。。

本题的思路相当直接，难点在于写出无bug的链表翻转操作。。

步骤：

得到linkedlist的node数量
通过linkedlist的node数量得到中点，以及中点前面的一点，中点前面的点指向nullptr，这样则成功把linkedlist分割成两个链表
将后面的链表翻转，具体操作可以参考我的这两篇博客: NO.206 和 NO.234
然后，将两个链表进行merge操作即可

详细代码如下：

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    void reorderList(ListNode* head) {
        
        if (head == nullptr or head->next == nullptr) {
            return;
        }
        // 1, count the number of nodes
        ListNode* tmp = head;
        int cnt = 0;
        while (tmp != nullptr) {
            ++cnt;
            tmp = tmp->next;
        }
        
        // 2, split the linkedlist
        ListNode* pre;
        tmp = head;
        for (int i = 0; i < (cnt + 1) / 2; ++i) {
            pre = tmp;
            tmp = tmp->next;
        }
        pre->next = nullptr;
        
        // 3, reverse the second half
        ListNode* new_node = new ListNode(0);
        ListNode* after = tmp->next;
        new_node->next = after;
        tmp->next = nullptr;
        while (after != nullptr) {
            after = after->next;
            new_node->next->next = tmp;
            tmp = new_node->next;
            new_node->next = after;
        }
        
        // 4, merge two linked list
        pre = head;
        ListNode* pre_next;
        ListNode* tmp_next;
        while (tmp != nullptr) {
            pre_next = pre->next;
            tmp_next = tmp->next;
            pre->next = tmp;
            tmp->next = pre_next;
            pre = pre_next;
            tmp = tmp_next;
        }
    }
};