148. Sort List

 

Sort a linked list in  O ( n  log  n ) time using constant space complexity.

Example 1:

Input: 4->2->1->3
Output: 1->2->3->4

Example 2:

Input: -1->5->3->4->0
Output: -1->0->3->4->5

 

常见排序方法有很多,插入排序,选择排序,堆排序,快速排序,冒泡排序,归并排序,桶排序等等。。它们的时间复杂度不尽相同,而这里题目限定了时间必须为O(nlgn),符合要求只有快速排序,归并排序,堆排序,而根据单链表的特点,最适于用归并排序。为啥呢?这是由于链表自身的特点决定的,由于不能通过坐标来直接访问元素,所以快排什么的可能不太容易实现(但是被评论区的大神们打脸,还是可以实现的),堆排序的话,如果让新建结点的话,还是可以考虑的,若只能交换结点,最好还是不要用。而归并排序(又称混合排序)因其可以利用递归来交换数字,天然适合链表这种结构。归并排序的核心是一个 merge() 函数,其主要是合并两个有序链表,这个在 LeetCode 中也有单独的题目 Merge Two Sorted Lists。由于两个链表是要有序的才能比较容易 merge,那么对于一个无序的链表,如何才能拆分成有序的两个链表呢?我们从简单来想,什么时候两个链表一定都是有序的?就是当两个链表各只有一个结点的时候,一定是有序的。而归并排序的核心其实是分治法 Divide and Conquer,就是将链表从中间断开,分成两部分,左右两边再分别调用排序的递归函数 sortList(),得到各自有序的链表后,再进行 merge(),这样整体就是有序的了。因为子链表的递归函数中还是会再次拆成两半,当拆到链表只有一个结点时,无法继续拆分了,而这正好满足了前面所说的“一个结点的时候一定是有序的”,这样就可以进行 merge 了。然后再回溯回去,每次得到的都是有序的链表,然后进行 merge,直到还原整个长度。这里将链表从中间断开的方法,采用的就是快慢指针,大家可能对快慢指针找链表中的环比较熟悉,其实找链表中的中点同样好使,因为快指针每次走两步,慢指针每次走一步,当快指针到达链表末尾时,慢指针正好走到中间位置,参见代码如下:

 

C++ 解法一:

class Solution {
public:
    ListNode* sortList(ListNode* head) {
        if (!head || !head->next) return head;
        ListNode *slow = head, *fast = head, *pre = head;
        while (fast && fast->next) {
            pre = slow;
            slow = slow->next;
            fast = fast->next->next;
        }
        pre->next = NULL;
        return merge(sortList(head), sortList(slow));
    }
    ListNode* merge(ListNode* l1, ListNode* l2) {
        ListNode *dummy = new ListNode(-1);
        ListNode *cur = dummy;
        while (l1 && l2) {
            if (l1->val < l2->val) {
                cur->next = l1;
                l1 = l1->next;
            } else {
                cur->next = l2;
                l2 = l2->next;
            }
            cur = cur->next;
        }
        if (l1) cur->next = l1;
        if (l2) cur->next = l2;
        return dummy->next;
    }
};

 

Java 解法一:

public class Solution {
    public ListNode sortList(ListNode head) {
        if (head == null || head.next == null) return head;
        ListNode slow = head, fast = head, pre = head;
        while (fast != null && fast.next != null) {
            pre = slow;
            slow = slow.next;
            fast = fast.next.next;
        }
        pre.next = null;
        return merge(sortList(head), sortList(slow));
    }
    public ListNode merge(ListNode l1, ListNode l2) {
        ListNode dummy = new ListNode(-1);
        ListNode cur = dummy;
        while (l1 != null && l2 != null) {
            if (l1.val < l2.val) {
                cur.next = l1;
                l1 = l1.next;
            } else {
                cur.next = l2;
                l2 = l2.next;
            }
            cur = cur.next;
        }
        if (l1 != null) cur.next = l1;
        if (l2 != null) cur.next = l2;
        return dummy.next;
    }
}

 

下面这种方法也是归并排序,而且在merge函数中也使用了递归,这样使代码更加简洁啦~

 

C++ 解法二:

class Solution {
public:
    ListNode* sortList(ListNode* head) {
        if (!head || !head->next) return head;
        ListNode *slow = head, *fast = head, *pre = head;
        while (fast && fast->next) {
            pre = slow;
            slow = slow->next;
            fast = fast->next->next;
        }
        pre->next = NULL;
        return merge(sortList(head), sortList(slow));
    }
    ListNode* merge(ListNode* l1, ListNode* l2) {
        if (!l1) return l2;
        if (!l2) return l1;
        if (l1->val < l2->val) {
            l1->next = merge(l1->next, l2);
            return l1;
        } else {
            l2->next = merge(l1, l2->next);
            return l2;
        }
    }
};

 

Java 解法二:

public class Solution {
    public ListNode sortList(ListNode head) {
        if (head == null || head.next == null) return head;
        ListNode slow = head, fast = head, pre = head;
        while (fast != null && fast.next != null) {
            pre = slow;
            slow = slow.next;
            fast = fast.next.next;
        }
        pre.next = null;
        return merge(sortList(head), sortList(slow));
    }
    public ListNode merge(ListNode l1, ListNode l2) {
        if (l1 == null) return l2;
        if (l2 == null) return l1;
        if (l1.val < l2.val) {
            l1.next = merge(l1.next, l2);
            return l1;
        } else {
            l2.next = merge(l1, l2.next);
            return l2;
        }
    }
}

 

Github 同步地址:

https://github.com/grandyang/leetcode/issues/148

 

类似题目:

Merge Two Sorted Lists

Sort Colors

Insertion Sort List

 

参考资料:

https://leetcode.com/problems/sort-list/description/

https://leetcode.com/problems/sort-list/discuss/46857/clean-and-short-merge-sort-solution-in-c

https://leetcode.com/problems/sort-list/discuss/46937/56ms-c-solutions-using-quicksort-with-explanations

https://leetcode.com/problems/sort-list/discuss/46772/i-have-a-pretty-good-mergesort-method-can-anyone-speed-up-the-run-time-or-reduce-the-memory-usage

 

LeetCode All in One 题目讲解汇总(持续更新中…)


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