[LeetCode] Sort List 链表排序

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

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。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;
        }
    }
}

本文转自博客园Grandyang的博客,原文链接:链表排序[LeetCode] Sort List ,如需转载请自行联系原博主。

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