148:Sort List【链表】【排序】

题目链接：https://leetcode.com/problems/sort-list/

/*题意：对链表进行排序*/

/**
 *思路：归并排序
 *      分治：将链表分成两段：用slow和fast指针，slow每次只走一步，fast每次
 *      走两步。当fast为空时，slow所在位置就是链表中间。
 *      合并：两个简单的链表的合并
 */
class Solution {
public:
    void show(ListNode *head) {
        while(head != NULL) {
            cout << head->val << " ";
            head = head->next;
        }
        cout << endl;
    }
    ListNode *Merge(ListNode *headA, ListNode *headB) {
        ListNode *root = new ListNode(0);
        ListNode *p = root;
        while(headA && headB) {
            if(headA->val <= headB->val) {
                p->next = headA;
                headA = headA->next;
            }
            else {
                p->next = headB;
                headB = headB->next;
            }
            p = p->next;
        }
        while(headA) {
            p->next = headA;
            p = p->next;
            headA = headA->next;
        }
        while(headB) {
            p->next = headB;
            p = p->next;
            headB = headB->next;
        }
        ListNode *res = root->next;
        delete root;
        return res;

    }
    ListNode *MergeSort(ListNode *head) {
        //只有一个结点
        if(head == NULL || head->next == NULL) return head;
        ListNode *slow = head;            //慢指针
        ListNode *fast = head->next->next;//快指针
        while(fast != NULL && fast->next != NULL) {
            slow = slow->next;
            fast = fast->next->next;
        }
        ListNode *headA = head;
        ListNode *headB = slow->next;
        slow->next = NULL;      //将链表断开
        headA = MergeSort(headA);
        headB = MergeSort(headB);
        return Merge(headA, headB);
    }
    ListNode* sortList(ListNode* head) {
        if(head == NULL || head->next == NULL)
            return head;
        return MergeSort(head);
    }
};