Reorder List (LeetCode)

Question:

https://oj.leetcode.com/problems/reorder-list/

 

解答：

inplace的解法，想到就是把后半段Linked List反转，然后逐个插入到上半段Linked List中。注意边界条件还有后半段是curr+1. (因为最后剩下不管是两个还是1个，都不需要处理，所以后半段要不比前半段少一个，要不少两个).

找到后半段List的头可以用两个指针，一个一次走一步，一个一次走两步。

class Solution {
public:
    void reorderList(ListNode *head) {
        if (!head)
            return;
        
        // Find the middle pointer, and reverse the rest list
        ListNode* curr = head;
        ListNode* curr2 = head;
        
        while (curr2 && curr2->next)
        {
            curr = curr->next;
            curr2 = curr2->next->next;
        }
        
        ListNode* rightHalfHead = curr->next;
        if (!rightHalfHead)
            return;

        curr->next = NULL;

        // reverse rightHalfHead
        curr2 = rightHalfHead;
        while (rightHalfHead->next)
        {
            ListNode* next = rightHalfHead->next->next;
            rightHalfHead->next->next = curr2;
            curr2 = rightHalfHead->next;
            
            rightHalfHead->next = next;
        }
        
        curr = head;
        while (curr2)
        {
            // insert curr2 after curr
            ListNode* next = curr2->next;
            curr2->next = curr->next;
            curr->next = curr2;
            curr2 = next;
            curr = curr->next->next;
        }
    }
};