[LeetCode] Reorder List

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

首先用一个vector存储所有的结点，然后将倒数第i个结点插入到第i个结点之后。

实现代码1

//Runtime:82 ms
#include <iostream>
#include <vector>
using namespace std;

struct ListNode
{
int val;
ListNode *next;
ListNode(int x) : val(x), next(NULL){}
};

class Solution {
public:
void reorderList(ListNode *head) {
if (head == NULL)
{
return;
}
ListNode *p = head;
vector<ListNode*> nodes;
while (p)
{
nodes.push_back(p);
p = p->next;
}

int left = 0;
int right = nodes.size() - 1;
while (left < right)
{
nodes[left]->next = nodes[right];
nodes[right--]->next = nodes[++left];
}
nodes[left]->next = NULL;
}
};
int main()
{
ListNode *p1 = new ListNode(1);
ListNode *p2 = new ListNode(2);
ListNode *p3 = new ListNode(3);
ListNode *p4 = new ListNode(4);
p1->next = p2;
p2->next = p3;
p3->next = p4;
Solution s;
s.reorderList(p1);

ListNode *p = p1;
while (p)
{
cout<<p->val<<endl;
p = p->next;
}
}

解题思路2

首先把链表从中间分成两部分，然后将后一部分链表反转，最后将两个链表合并。

实现代码2

//Runtime:98 ms
class Solution {
public:
void reorderList(ListNode *head) {
if (head == NULL)
{
return;
}
ListNode* ps = head;
ListNode* pf = head;
while (pf->next && pf->next->next)
{
ps = ps->next;
pf = pf->next->next;
}

ListNode *p2 = ps->next;
ps->next = NULL;
p2 = reverseList(p2);
head = mergeList(head, p2);
}

ListNode* mergeList(ListNode *p1, ListNode *p2)
{
ListNode *result = p1;
ListNode *temp;
while (p2)
{
temp = p2;
p2 = p2->next;

temp->next = p1->next;
p1->next = temp;
p1 = p1->next->next;
}

return result;
}

ListNode* reverseList(ListNode *head){
if (head == NULL)
{
return NULL;
}
ListNode *p = head->next;
head->next = NULL;
while (p)
{
ListNode *temp = p;
p = p->next;

temp->next = head;
head = temp;
}

return head;
}
};