LeetCode 143. Reorder List

题目描述：

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}.

解题思路：

这是一个很常规的题目， 拿到这道题目只需要将链表分割成前后两段，可以借助快慢指针的做法， 然后将后面一个链表逆置， 再按照指定的次序排序就可以了。

下面是代码：

/**
* 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) {
// 先找到链表的中心位置， 切开
ListNode myhead(0);
myhead.next = head;
ListNode * p1 = &myhead, *p2 = &myhead;
while (p1 && p1->next){
p1 = p1->next->next;
p2 = p2->next;
}
ListNode * pNext = p2->next;

// 将后面一部分链表切开
ListNode myhead2(0);
p2->next = nullptr;
myhead2.next = pNext;

// 将链表2 逆置
ListNode * pCur = myhead2.next;
ListNode * pPre = nullptr;
while (pCur){
ListNode * pNext = pCur->next;
pCur->next = pPre;
pPre = pCur;
pCur = pNext;
}
myhead2.next = pPre;

// 归并排序, 因为p1 >= p2
p1 = myhead.next;
p2 = myhead2.next;
ListNode res(0);
ListNode * pres = &res;
while (p1){
pres->next = p1;
p1 = p1->next;
pres = pres->next;
pres->next = nullptr;

if (p2){
pres->next = p2;
p2 = p2->next;
pres = pres->next;
pres->next = nullptr;
}
}
}
};