148. 排序链表

给你链表的头结点 head ，请将其按 升序 排列并返回 排序后的链表 。

输入：head = [4,2,1,3]
输出：[1,2,3,4]

> 思路1：
采用二路归并排序，分为递归和迭代 两种方法。

递归采用快慢指针，找到链表的中点，断开，然后对两条链表进行二路归并

class Solution {
public:
    ListNode* sortList(ListNode* head) {
        if(!head || !head->next) return head;
        ListNode* pre = head, *slow = head, *fast = head;
        while(fast && fast->next) {
            pre = slow;
            slow = slow->next;
            fast = fast->next->next;
        }
        pre->next = nullptr;
        return mergeTwoList(sortList(head), sortList(slow));
    }
    ListNode* mergeTwoList(ListNode* h1, ListNode* h2) {
        if(!h1) return h2;
        if(!h2) return h1;
        if(h1->val < h2->val) {
            h1->next = mergeTwoList(h1->next, h2);
            return h1;
        }
        else {
            h2->next = mergeTwoList(h1, h2->next);
            return h2;
        }
    }
};

> 思路2：
迭代的话，我们采用自下向顶的方法，如图

代码中的tail节点用来连接被切断的节点

class Solution {
public:
    ListNode* sortList(ListNode* head) {
        ListNode dummyHead(0);
        dummyHead.next = head;
        auto p = head;
        int length = 0;
        while (p) {
            ++length;
            p = p->next;
        }
        
        for (int size = 1; size < length; size <<= 1) {
            auto cur = dummyHead.next;
            auto tail = &dummyHead;
            
            while (cur) {
                auto left = cur;
                auto right = cut(left, size); // left->@->@ right->@->@->@...
                cur = cut(right, size); // left->@->@ right->@->@  cur->@->...
                
                tail->next = merge(left, right);
                while (tail->next) {
                    tail = tail->next;
                }
            }
        }
        return dummyHead.next;
    }
    
    ListNode* cut(ListNode* head, int n) {
        auto p = head;
        while (--n && p) {
            p = p->next;
        }
        
        if (!p) return nullptr;
        
        auto next = p->next;
        p->next = nullptr;
        return next;
    }
    
    ListNode* merge(ListNode* l1, ListNode* l2) {
        ListNode dummyHead(0);
        auto p = &dummyHead;
        while (l1 && l2) {
            if (l1->val < l2->val) {
                p->next = l1;
                p = l1;
                l1 = l1->next;       
            } else {
                p->next = l2;
                p = l2;
                l2 = l2->next;
            }
        }
        p->next = l1 ? l1 : l2;
        return dummyHead.next;
    }
};