[LeetCode] 148. Sort List 链表排序

2021年09月15日 阅读数:1
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Sort a linked list in O(n log n) time using constant space complexity.html

Example 1:java

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

Example 2:node

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

给一个链表排序,要求Time: O(nlogn), constant space complexity.python

解法:1. 把链表从中间分开Break the list to two in the middle.  2. 递归排序两个子链表Recursively sort the two sub lists.  3. 合并子链表Merge the two sub lists. post

解法:归并排序。因为有时间和空间复杂度的要求。把链表从中间分开,递归下去,都最后两个node时开始合并,返回上一层继续合并,直到结束。找中间点的方法能够用快慢指针,快指针走2步,慢指针走1步。也能够求出链表长度,再分开链表url

Java:spa

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;
    }
}

Java:指针

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;
        }
    }
}

Python:orm

class ListNode:
    def __init__(self, x):
        self.val = x
        self.next = None

    def __repr__(self):
        if self:
            return "{} -> {}".format(self.val, repr(self.next))

class Solution:
    # @param head, a ListNode
    # @return a ListNode
    def sortList(self, head):
        if head == None or head.next == None:
            return head

        fast, slow, prev = head, head, None
        while fast != None and fast.next != None:
            prev, fast, slow = slow, fast.next.next, slow.next
        prev.next = None

        sorted_l1 = self.sortList(head)
        sorted_l2 = self.sortList(slow)

        return self.mergeTwoLists(sorted_l1, sorted_l2)

    def mergeTwoLists(self, l1, l2):
        dummy = ListNode(0)

        cur = dummy
        while l1 != None and l2 != None:
            if l1.val <= l2.val:
                cur.next, cur, l1 = l1, l1, l1.next
            else:
                cur.next, cur, l2 = l2, l2, l2.next

        if l1 != None:
            cur.next = l1
        if l2 != None:
            cur.next = l2

        return dummy.next

if __name__ == "__main__":
    head = ListNode(3)
    head.next = ListNode(4)
    head.next.next = ListNode(1)
    head.next.next.next= ListNode(2)
    print(Solution().sortList(head))  

C++:htm

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;
    }
};

C++:

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;
        }
    }
};

  

  

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