leetcode148——Sort List

题目大意：时间复杂度O(nlogn)，空间复杂度O(1)对链表排序

分析：满足时间空间复杂度的排序方法——非递归归并（bottom-to-up）。1个和1个合并=>2个和2个合并=>4个和4个合并，自底向上完成归并排序。

代码：

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode* sortList(ListNode* head) {
        ListNode* dummy = new ListNode(0);
        dummy->next = head;
        ListNode *h = head, *h1, *h2, *pre;
        int length = 0, unit = 1;
        while (h) {
            h = h->next;
            length++;
        }
        while (unit < length) {
            pre = dummy;  //用pre来牵引合并的过程
            h = dummy->next; //h指向两组要合并的unit的头节点
            while (h) {
                //接下来要获取两组的头h1和h2，还有两组的长度c1和c2
                int i = unit;
                h1 = h;
                while (i > 0 && h) {
                    h = h->next;
                    i--;
                }
                if (i > 0) break; //剩下的结点个数不够unit，无法做两组unit合并
                i = unit;
                h2 = h;
                while (i > 0 && h) {
                    h = h->next;
                    i--;
                }
                int c1 = unit, c2 = unit - i; //即将合并的两段的长度
                while (c1 > 0 && c2 > 0) {  //利用pre牵引从小到大的结点
                    if (h1->val < h2->val) {
                        pre->next = h1;
                        h1 = h1->next;
                        c1--;
                    } 
                    else {
                        pre->next = h2;
                        h2 = h2->next;
                        c2--;
                    }
                    pre = pre->next;
                }
                pre->next = c1 == 0 ? h2 : h1;
                while (c1 > 0 || c2 > 0) {  //pre牵引到合并后的两组unit的尾结点
                    pre = pre->next;
                    c1--;
                    c2--;
                }
                pre->next = h;  //合并后的两组unit的尾结点与下一组头节点相连
            }
            unit *= 2;
        }
        return dummy->next;
    }
};