前言
华为三道题，100+200+300，100及格，大家做对第一题就好了，祝大家全都有心仪的offer，不要慌，不要焦虑
一、完美排列——玩具（全A）（注意：题目中说：如果不是完美排列，则输出0，没注意这种情况的应该A0.6或0.7）
代码：暴力就完事了
package huawei0909;
import java.util.Scanner;
/**
* Created by IntelliJ IDEA.
*
* @Author:
* @Email:
* @Date: 2020/9/9
* @Time: 19:04
* @Version: 1.0
* @Description: Description
*/
public class First {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
int K = sc.nextInt();
int[] perArr = new int[K];
int[] perArr1 = new int[K];
for (int i = 0; i < K; i++)
perArr[i] = sc.nextInt();
for (int i = 0; i < K; i++)
perArr1[i] = sc.nextInt();
int n = sc.nextInt();
int[] arr = new int[n];
int[] arr1 = new int[n];
for (int i = 0; i < n; i++)
arr[i] = sc.nextInt();
for (int i = 0; i < n; i++)
arr1[i] = sc.nextInt();
sc.close();
/*if (n<K){ //必须判断，不然A0.6或0.7，因为题目中说：如果不是完美排列，则输出0，详情看下面的System.out.println(i + 1);
System.out.println(0);
return;
}*/
for (int i = 0; i < n; i++) {
if (arr[i] == perArr[0] && arr1[i] == perArr1[0] && i + K - 1 < n && arr[i + K - 1] == perArr[K - 1] && arr1[i + K - 1] == perArr1[K - 1]) {
boolean flag = true;
int index = i;
for (int j = 1; j < K - 1; j++) {
index++;
if (!(arr[index] == perArr[j] && arr1[index] == perArr1[j])) {
flag = false;
break;
}
}
if (flag) { //输出可能为0，如果没考虑到，则A0.6或0.7，因为题目中说：如果不是完美排列，则输出0
System.out.println(i + 1);
return;
}
}
}
System.out.println(0); //必须有，不然A0.6或0.7，因为题目中说：如果不是完美排列，则输出0，详情看下面的System.out.println(i + 1);
}
}
二、最长的水沟（全A）
package huawei0909;
import java.util.Scanner;
/**
* Created by IntelliJ IDEA.
*
* @Author:
* @Email:
* @Date: 2020/9/9
* @Time: 19:36
* @Version: 1.0
* @Description: Description
*/
public class Second {
public static int[][] matrix;
public static int[][] dp;
public static int[][] k = {{1, 0}, {0, 1}, {-1, 0}, {0, -1}};
public static int n, m, ans;
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
n = sc.nextInt();
m = sc.nextInt();
matrix = new int[n + 1][m + 1];
dp = new int[n + 1][m + 1];
for (int i = 1; i <= n; i++)
for (int j = 1; j <= m; j++)
matrix[i][j] = sc.nextInt();
for (int i = 1; i <= n; i++)
for (int j = 1; j <= m; j++)
ans = Math.max(ans, dfs(i, j));
System.out.println(ans + 1);
}
public static int dfs(int x, int y) {
if (dp[x][y] != 0)
return dp[x][y];
for (int i = 0; i <= 3; i++) {
int tx = x + k[i][0];
int ty = y + k[i][1];
if (!(tx < 1 || ty < 1 || tx > n || ty > m || matrix[tx][ty] >= matrix[x][y]))
dp[x][y] = Math.max(dp[x][y], 1 + dfs(tx, ty));
}
return dp[x][y];
}
}