leetcode143. 重排链表 Reorder List

给定一个单链表 L：L0→L1→…→Ln-1→Ln ，
将其重新排列后变为： L0→Ln→L1→Ln-1→L2→Ln-2→…

你不能只是单纯的改变节点内部的值，而是需要实际的进行节点交换。

示例 1:

给定链表 1->2->3->4, 重新排列为 1->4->2->3.
示例 2:

给定链表 1->2->3->4->5, 重新排列为 1->5->2->4->3.

1.快慢指针找到中点 2.拆成两个链表 3.遍历两个链表，后面的塞到前面的“缝隙里”

    void reorderList(ListNode* head) {
        
        if(head==NULL || head->next == NULL)
            return;
        
        //快慢指针分出两段
        ListNode *slow = head,*fast = head;
        
        while(fast->next && fast->next->next){
            
            slow = slow->next;
            fast = fast->next->next;
        }
        
        
        //后端反转
        ListNode *needReverser = slow->next;
        slow->next = NULL;
        needReverser = reverse(needReverser);
        
        //插入前端缝隙
        ListNode *cur = head;
        while(cur && needReverser){
            ListNode *curSecond = needReverser;
            needReverser = needReverser->next;
            ListNode *nextCur = cur->next;
            curSecond->next = cur->next;
            cur->next = curSecond;
            
            cur = nextCur;
        }
        
    }
    
    ListNode *reverse(ListNode *head){
        
        ListNode *p1 = NULL;
        ListNode *p2 = head;
        ListNode *p3 = p2;
        
        while(p2){
            p3 = p2->next;
            p2->next = p1;
            p1 = p2;
            p2 = p3;            
        }
        
        return p1;
        
        
    }