Sort List & Insertion Sort List

148. Sort List

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

题目大意：给定一个链表，要求以O(nlogn)时间复杂度和O(1)空间复杂度将链表排序。

思路：使用归并排序

算法步骤：（1）使用快慢指针将链表分为两部分，如果链表长度小于2，则不划分，直接返回链表；（2）将得到的两个链表分别排序；（3）合并两个有序链表。

算法复杂度：时间复杂度O(nlog(n))，空间复杂度O(1)

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) {
        if (!head || !head->next)
            return head;
        ListNode *fast = head, *slow = head;
        while (fast->next && fast->next->next) {
            fast = fast->next->next;
            slow = slow->next;
        }
        fast = slow->next;
        slow->next = nullptr;
        
        ListNode *l1 = sortList(head);
        ListNode *l2 = sortList(fast);
        
        return mergeList(l1, l2);
    }
    ListNode *mergeList(ListNode *l1, ListNode *l2) {
        ListNode res(0), *pres = &res;
        while (l1 && l2) {
            if (l1->val <= l2->val) {
                pres->next = l1;
                l1 = l1->next;
            }
            else {
                pres->next = l2;
                l2 = l2->next;
            }
            pres = pres->next;
        }
        pres->next = l1 ? l1 : l2;
        return res.next;
    }
};

评测系统上运行结果：

147. Insertion Sort List

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

题目大意：给定一个链表，使用插入排序算法对链表进行排序

算法步骤：（1）while循环找到插入点；（2）将当前节点插入插入点。

算法复杂度：时间复杂度O(n^2)，空间复杂度O(1)

代码：

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode* insertionSortList(ListNode* head) {
        if (!head || !head->next)
            return head;
        ListNode pstart(0), *pcurr = head->next, *ppre = head;
        pstart.next = head;
        while(pcurr) {
            ListNode *p = &pstart;
            while (p->next != pcurr && p->next->val <= pcurr->val)
                    p = p->next;
            if (p->next == pcurr) {
                ppre = ppre->next;
                pcurr = pcurr->next;
            } else {
                ppre->next = pcurr->next;
                pcurr->next = p->next;
                p->next = pcurr;
                pcurr = ppre->next;
            }
        }
        return pstart.next;
    }
};

评测系统上运行结果：