LeetCode143：Reorder List

题目：

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：

1，先利用快慢指针找到链表中间节点

2，将链表后半部分进行反转

3，将链表前半部分与反转后的后半部分进行合并

实现代码：

#include <iostream>
using namespace std;

/*
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}.
*/
struct ListNode {
int val;
ListNode *next;
ListNode(int x) : val(x), next(NULL) {}
};
void addNode(ListNode* &head, int val)
{
ListNode *node = new ListNode(val);
if(head == NULL)
{
head = node;
}
else
{
node->next = head;
head = node;
}
}
void printList(ListNode *head)
{
while(head)
{
cout<<head->val<<" ";
head = head->next;
}
}
class Solution {
public:
void reorderList(ListNode *head) {
if(head == NULL || head->next == NULL)
return ;
ListNode *quick = head;
ListNode *slow = head;
while(quick->next &&quick->next->next)//采用快慢指针查找链表中间节点，快指针走两步，慢指针走一步
{
quick = quick->next->next;
slow = slow->next;
}
quick = slow;
slow = slow->next;
quick->next = NULL;
reverseList(slow);//将后半部分进行反转

quick = head;
while(quick && slow)//将前半部分与反转后的后半部分进行合并
{
ListNode *t = slow->next;
slow->next = quick->next;
quick->next = slow;
slow = t;
quick = quick->next->next;
}

}
void reverseList(ListNode* &head)//采用头插法进行链表反转
{
if(head == NULL || head->next == NULL)
return ;
ListNode *p = head->next;
head->next = NULL;
while(p)
{
ListNode *t = p->next;
p->next = head;
head = p;
p = t;
}
}
};
int main(void)
{
ListNode *head = new ListNode(6);
addNode(head, 5);
addNode(head, 4);
addNode(head, 3);
addNode(head, 2);
addNode(head, 1);
addNode(head, 0);
printList(head);
cout<<endl;

Solution solution;
solution.reorderList(head);
printList(head);

return 0;
}