算法内容
本文件实现 LeetCode 148「排序链表」。核心思路是对单链表做归并排序，利用链表切分和有序链表合并完成排序。
split 使用快慢指针找到链表中点，并断开为左右两段；sortList 递归排序左右子链表；merge 使用虚拟头节点把两个有序链表合并。归并排序适合链表结构，因为合并时只需要调整指针，不需要随机访问。时间复杂度为 O(n log n)，递归栈空间复杂度为 O(log n)。
算法实现
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#include <vector>
#include <iostream>
#include <stack>
#include <queue>
#include <algorithm>
#include <unordered_map>
struct ListNode
{
int val;
ListNode *next;
ListNode() : val(0), next(nullptr) {}
ListNode(int x) : val(x), next(nullptr) {}
ListNode(int x, ListNode *next) : val(x), next(next) {}
};
using namespace std;
class Solution
{
public:
ListNode *sortList(ListNode *head)
{
ListNode *left, *right;
int len = split(head, &left, &right);
if (len > 2)
{
left = sortList(left);
right = sortList(right);
}
return merge(left, right);
}
ListNode *merge(ListNode *left, ListNode *right)
{
ListNode *head = new ListNode(0), *tmp;
tmp = head;
while (left != nullptr && right != nullptr)
{
if (left->val < right->val)
{
tmp->next = left;
left = left->next;
}
else
{
tmp->next = right;
right = right->next;
}
tmp = tmp->next;
}
if (left != nullptr)
{
tmp->next = left;
}
if (right != nullptr)
{
tmp->next = right;
}
tmp = head->next;
return tmp;
}
int split(ListNode *head, ListNode **left, ListNode **right)
{
ListNode *v = new ListNode(0);
v->next = head;
ListNode *slow = v, *fast = v;
int len = 0;
while (fast->next != nullptr)
{
slow = slow->next;
fast = fast->next;
len++;
if (fast->next != nullptr)
{
fast = fast->next;
len++;
}
}
*left = head;
*right = slow->next;
slow->next = nullptr;
return len;
}
};
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