2021字節跳動秋招校招算法面試熱門真題精講， 內含7種語言答案

排序鏈表

1.題目描述

<p>在 O(n log n) 時間複雜度和常數級空間複雜度下，對鏈表進行排序。</p><br/><p>示例 1:</p><br/><pre>輸入: 4->2->1->3<br/>輸出: 1->2->3->4<br/></pre><br/><p>示例 2:</p><br/><pre>輸入: -1->5->3->4->0<br/>輸出: -1->0->3->4->5</pre><br/>

c答案

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     struct ListNode *next;
 * };
 */

//head不參與排序！
void mergeSortList(struct ListNode* head) {
    struct ListNode* slow, *fast,*pre;
    if (NULL == head)
        return;    //無效輸入
    else{
        struct ListNode R;
        slow = fast = head;
        do {
            if ((fast = fast->next) != NULL) {
                fast = fast->next;
                slow = slow->next;
            }
        } while (fast);
        if (NULL == slow->next)return;    //只有0或1個元素
        R.next = slow->next;
        slow->next = NULL;    //一分爲二
        mergeSortList(head);    //使左半有序
        mergeSortList(&R);        //使右半有序
        slow = head->next;    fast = R.next;    //兩邊準備進行歸併
    }
    pre = head;    //尾插法（是slow的前驅）
    do {
        if (fast->val < slow->val) {    //右邊比左邊小
            pre = pre->next = fast;        //原pre->next==slow，因此不會丟鏈
            fast = fast->next;
            pre->next = slow;            //還原pre是slow的前驅
            if (fast == NULL) 
                break;        //左邊的pre原本就接着slow，無需處理
        }
        else {    //左邊小
            pre = slow;
            slow = slow->next;
            if (slow == NULL) {
                pre->next = fast;        //剩餘的右邊接到左邊
                break;
            }
        }
    } while (true);
}

struct ListNode* sortList(struct ListNode* head) {
    struct ListNode Head; Head.next = head;    //真正頭結點，不參與排序
    mergeSortList(&Head);
    return Head.next;
}

c++答案

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode * merge(ListNode* left, ListNode* right)
    {
        ListNode* newhead = NULL, *pre=NULL;
        while (left && right)
        {
            if (left->val < right->val)
            {
                if (!newhead)
                {
                    newhead = pre = left;
                    left = left->next;
                }
                else
                {
                    pre->next = left;
                    pre = pre->next;
                    left = left->next;
                }
            }
            else
            {
                if (!newhead)
                {
                    newhead = pre = right;
                    right = right->next;
                }
                else
                {
                    pre->next = right;
                    pre = pre->next;
                    right = right->next;
                }
            }
        }
        while (left)
        {
            pre->next = left;
            left = left->next;
            pre = pre->next;
        }
        while (right)
        {
            pre->next = right;
            right = right->next;
            pre = pre->next;
        }
        pre->next = NULL;
        return newhead;
    }

    ListNode* sortList(ListNode* head) {
        if (head == NULL || head->next == NULL)
            return head;
        else
        {
            ListNode* slow = head, *fast = head->next;
            while (fast != NULL && fast->next != NULL)
            {
                slow = slow->next;
                fast = fast->next->next;
            }
            ListNode* temp = slow->next;
            slow->next = NULL;
            ListNode* l = sortList(head);
            ListNode* r = sortList(temp);
            return merge(l, r);
        }
    }
};

java答案

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
class Solution {
    public ListNode sortList(ListNode head) {
        if (head == null || head.next == null) return head;
        ListNode fast = head;
        ListNode slow = head;
        // 快慢指針尋找鏈表的中間節點
        while (fast.next != null && fast.next.next != null) {
            fast = fast.next.next;
            slow = slow.next;
        }
        // 把鏈表分爲兩段
        ListNode l1 = head;
        ListNode l2 = slow.next;
        slow.next = null;

        // Merge Sort
        l1 = sortList(l1);
        l2 = sortList(l2);
        return merge(l1, l2);
    }

    // 歸併兩個已經排好序的鏈表
    private ListNode merge(ListNode l1, ListNode l2) {
        if (l1 == null) return l2;
        if (l2 == null) return l1;

        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;
        else cur.next = l2;
        return dummy.next;
    }
}

JavaScript答案

/**
 * Definition for singly-linked list.
 * function ListNode(val) {
 *     this.val = val;
 *     this.next = null;
 * }
 */
/**
 * @param {ListNode} head
 * @return {ListNode}
 */
var merge = function(l1,l2) {
    var l = new ListNode();
    var p = l;
    while(l1 && l2) {
        if(l1.val < l2.val) {
            p.next = l1;
            l1 = l1.next
        } else {
            p.next = l2;
            l2 = l2.next
        }
        p = p.next
    }
    if(l1) {
        p.next = l1;
    }
    if(l2) {
        p.next = l2;
    }
    return l.next;
}
var sortList = function(head) {
    if(head === null || head.next === null) {
       return head;
    }
    var slow = head;
    var fast = head;
    var prev = null;
    while(fast && fast.next) {
        prev = slow
        fast = fast.next.next;
        slow = slow.next
    }
    prev.next = null;
    return merge(sortList(head), sortList(slow));
};

c#答案

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     public int val;
 *     public ListNode next;
 *     public ListNode(int x) { val = x; }
 * }
 */
public class Solution {
    public ListNode SortList(ListNode head) {
        if(head == null) return head;    
        ListNode now = head;
            List<int> list = new List<int>();
            while(now != null)
            {
                list.Add(now.val);
                now = now.next;
            }
            list.Sort();
            int length = list.Count;
            ListNode result = new ListNode(list[0]);
            now = result;
            for(int i=1;i<length;i++)
            {
                now.next = new ListNode(list[i]);
                now = now.next;
            }
            return result;
    }
}

python2.x答案

# Definition for singly-linked list.
# class ListNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.next = None

class Solution(object):
    def sortList(self, head):
        """
        :type head: ListNode
        :rtype: ListNode
        """
        node = head
        res=[]
        while node:
            res.append(node.val)
            node = node.next
        res.sort()
        if not res:
            return None
        newhead = ListNode(res[0])
        cur_Node = newhead
        for i,v in enumerate(res):
            if i == 0:
                continue
            obj = ListNode(v)
            cur_Node.next=obj
            cur_Node=obj
        return newhead

python3.x答案

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, x):
#         self.val = x
#         self.next = None

class Solution:
    def sortList(self, head):
        """
        :type head: ListNode
        :rtype: ListNode
        """
        if not head: return None 
        res = []
        while head:
            res.append(head)
            head = head.next
        res.sort(key = lambda ele: ele.val)
        temp = res[0]
        for i in range(1,len(res)):
            temp.next = res[i]
            temp = res[i]
        res[-1].next = None 
        return res[0]

go答案

/**
 * Definition for singly-linked list.
 * type ListNode struct {
 *     Val int
 *     Next *ListNode
 * }
 */
func sortList(head *ListNode) *ListNode {
    if head == nil || head.Next == nil {
        return head
    }
    
    middle := split(head)
    
    return merge(sortList(head), sortList(middle))
}

// 雙指針切分list
func split(head *ListNode) *ListNode {
    slow, fast := head, head
    var tail *ListNode
    
    for fast != nil && fast.Next != nil {
        tail = slow
        slow = slow.Next
        fast = fast.Next.Next
    }
    
    tail.Next = nil
    return slow
}

func merge(left ,right *ListNode) *ListNode {
    var head, cur, pre *ListNode
    for left != nil && right != nil {
        if left.Val < right.Val {
            cur = left
            left = left.Next
        } else {
            cur = right
            right = right.Next
        }
        // 生成head
        if head == nil {
            head = cur
        } else {
            pre.Next = cur
        }
        pre = cur
    }
    
    if left == nil {
        pre.Next = right
    } else {
        pre.Next = left
    }
    
    return head
}

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