算法笔试面试1

Posted 2020-12-08 trevo

蛇形矩阵

题目链接： https://www.acwing.com/activity/content/problem/content/1898/1/

思路

设计一个偏移量，4个方向，走不通了就改变方向

实现

#include <iostream>
#include <algorithm>
#include <cstring>
using namespace std;

const int N = 100;
int n, m;
int res[N][N];

int main()
{
    cin >> n >> m;
    int dx[] = {0, 1, 0, -1}, dy[] = {1, 0, -1, 0};
    for(int x = 0, y = 0, k = 1, d = 0; k <= n * m; k++)
    {
        res[x][y] = k;
        int a = x + dx[d], b = y + dy[d];
        if(a < 0 || a >= n || b < 0 || b >= m || res[a][b]) 
        {
            d = (d + 1) % 4;
            a = x + dx[d], b = y + dy[d];
        }
        x = a, y = b;
    }
    for(int i =  0; i < n; i++)
    {
        for(int j = 0; j < m; j++)
            cout << res[i][j] << " ";
        cout << endl;
    }
    return 0;
}

单链表快速排序

题目链接：https://www.acwing.com/activity/content/problem/content/1899/1/

思路

创建三个指针，left,mid,right，分别是比基准小，等于基准，大于基准的数字，递归在对left，right两个链表进行排序
最后将三个链表拼接起来即可，记得最后释放空间。
链表的快速排序算法是稳定排序。

实现

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    //得到尾节点
    ListNode* get_tail(ListNode* head){
        while(head -> next) head = head -> next;
        return head;
    }
    ListNode* quickSortList(ListNode* head) {
        if(!head || !head->next) return head;
        auto left = new ListNode(-1), mid = new ListNode(-1), right = new ListNode(-1);
        auto ltail = left, mtail = mid, rtail = right;
        
        int val = head -> val;
        for(auto p = head; p; p = p -> next)
        {
            if(p -> val < val) ltail = ltail -> next = p; 
            else if(p -> val == val) mtail  = mtail -> next = p;
            else rtail = rtail -> next = p;
        }
        ltail -> next = mtail -> next = rtail -> next = NULL;
        left -> next = quickSortList(left -> next);
        right -> next = quickSortList(right -> next);
        
        //拼接三个链表
        get_tail(left) -> next = mid -> next;
        get_tail(left) -> next = right -> next;
        
        auto p = left -> next;
        delete left;
        delete mid;
        delete right;
        return p;
    }
};

链表的归并排序

迭代

递归

寻找矩阵的极小值

题目链接：https://www.acwing.com/activity/content/problem/content/1900/1/

思路

leetcode peek 162

实现

// Forward declaration of queryAPI.
// int query(int x, int y);
// return int means matrix[x][y].

class Solution {
public:
    vector<int> getMinimumValue(int n) {
        typedef long long LL;
        const LL INF = 1e15;
        
        int l = 0, r = n - 1;
        while(l < r)
        {
            int mid = (l + r) >> 1; //中点
            
            int k;
            LL val = INF;
            //求这一列的最小值
            for(int i = 0; i < n; i++)
            {
                int t = query(i, mid);
                if(t < val)
                {
                    val = t;
                    k = i;
                }
            }
            LL left = mid ? query(k, mid - 1) : INF;
            LL right = mid + 1 < n ? query(k, mid + 1) : INF;
            
            if(val < left && val < right) return {k, mid};
            if(left < val) r = mid - 1;
            else l = mid + 1;
        }
         
        int k;
        LL val = INF;
        for(int i = 0; i < n; i++)
        {
            int t = query(i, r);
            if(t < val)
            {
                val = t;
                k = i;
            }
        }
        return {k, r};
    }
};

鸡蛋的硬度

题目链接：https://www.acwing.com/activity/content/problem/content/1901/1/

思路

实现

#include <iostream>
#include <algorithm>
#include <cstring>
using namespace std;

const int N = 110, M = 11;

int n,m;
int f[N][M];

int main()
{
    while(cin >> n >> m)
    {
        for(int i = 1; i <= n; i++) f[i][1] = i;
        
        for(int i = 1; i <= m; i++) f[1][i] = 1;
        
        
        for(int i = 2; i <= n; i++)
            for(int j = 2; j <= m; j++)
            {
                f[i][j] = f[i][j - 1];
                for(int k = 1; k <= i; k++)
                    f[i][j] = min(f[i][j], max(f[k - 1][j - 1], f[i - k][j]) + 1);
            }
        cout << f[n][m] << endl;
    }
    return 0;
}