leetcode143. Reorder List

Posted 2020-06-15

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}.

这一题涉及到了 单链表的中点查找、部分反序、合并 三种操作。

搞了很久，头晕。

1.查找采用的是两倍距离方法，两个指针采用步长差2倍的方式向后移动。这个方法在链表的遍历中，很受用。链表于数组不同，没有索引，所以要定位到某一个节点，必须要经过遍历。于计算总长度，然后再按照位置查找来说，次方法很高效。

2.反序采用三个指针辅助完成。

3.合并：略过。

/*******
    step 1: finding the middle node
    step 2: reverse the the latter half
    step 3: merge two part
     */
    void reorderList(ListNode* head)
    {
        if (head == NULL || head -> next == NULL) {
            return ;
        }
        ListNode *p1, *p2, *p3, *head2;
        p1 = head;
        p2 = p1 -> next;
        while (p2 && p2 -> next) {
            p1 = p1 -> next;
            p2 = p2 -> next -> next;
        }
        
        head2 = p1 -> next;
        p1 -> next = NULL;
        // above code find head2 node which is the mid node
        p2 = head2 -> next;
        head2 -> next = NULL;
        while (p2) {
            p1 = p2 -> next;
            p2 -> next = head2;
            head2 = p2;
            p2 = p1;
        }
        // above code reverse the half part and head2 is the end node
        for (p1 = head, p2 =head2; p1; ) {
            auto tmp = p1 -> next;
            p1 = p1 -> next = p2;
            p2 = tmp;
        }
　　　　// above code merge the two part to one
    }