阿里笔试8.19

第一题：两个数组按照从前端依次选取的方式是否能拼接成相应数组。递归实现代码如下所示：

import java.util.Scanner;

public class Day_2020_8_19 {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        // 输入测试数据
        int T = sc.nextInt();
        for (int i = 0; i < T; i++) {
            // 输入数组个数
            int n = sc.nextInt();
            int m = sc.nextInt();
            int[] a = new int[n];
            int[] b = new int[m];
            // 输入数组a
            for (int j = 0; j < n; j++) {
                a[j] = sc.nextInt();
            }
            // 输入数组b
            for (int k = 0; k < m; k++) {
                b[k] = sc.nextInt();
            }
            int[] c = new int[n+m];
            // 处理逻辑
            boolean possible = isPossible(a, 0, b, 0, c, 0);
            if (possible) {
                System.out.println("possible");
            } else {
                System.out.println("no possible");
            }
        }
    }

    public static boolean isPossible(int[] a, int ta, int[] b, int tb,
                                     int[] c, int tc) {
        int aLen = a.length;
        int bLen = b.length;
        int cLen = c.length;
        if (ta==aLen) {
            while (tc!=cLen && tb!=bLen) {
                if (c[tc]!=b[tb]) {
                    return false;
                }
            }
        } else if (tb==bLen) {
            while (tc!=cLen && ta!=aLen) {
                if (c[tc]!=a[ta]) {
                    return false;
                }
            }
        } else {
            if (a[ta]==c[tc] && b[tb]==c[tc]) {
                return isPossible(a, ta+1, b, tb, c, tc+1)
                        || isPossible(a, ta, b, tb+1, c, tc+1);
            } else if (a[ta]==c[tc]) {
                return isPossible(a, ta+1, b, tb, c, tc+1);
            } else if (b[tb]==c[tc]) {
                return isPossible(a, ta, b, tb+1, c, tc+1);
            } else {
                return false;
            }
        }
        return true;
    }
}

第二题：求k个连续队员中最大体力和的最小个人体力，代码如下所示：

public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        while (sc.hasNext()) {
            // 输入正整数n（数组个数）
            int n = sc.nextInt();
            // 输入对应每个队员能力值
            int[] a = new int[n];
            for (int i = 0; i < n; i++) {
                a[i] = sc.nextInt();
            }

            for (int k=1; k<=a.length; k++) {
                int i = maxMin(a, k);
                System.out.print(i + " ");
            }
        }
    }

    public static int maxMin(int[] a, int k) {
        int length = a.length;
        int max = Integer.MIN_VALUE;
        int index = 0;
        for (int i = 0; i < length-k+1; i++) {
            int result = 0;
            for (int j=i; j< i+k+1; j++) {
                result = result + a[j];
            }
            if (result>max) {
                max = result;
                index = i;
            }
        }
        int min = a[index];
        for (int i= index; i< index+k+1; i++){
            if (a[i] < min) {
                min = a[i];
            }
        }
        return min;
    }
}

总结：这两道题是基础题，但是也是非常容易出错的题，因为你可能考虑的不全面，所以我贴出来和大家交流一下。有不对的地方请指正。