中兴秋招笔试

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中兴秋招笔试编程题

两个严格递增数组的逆序对

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public class Main {
    public static void main(String[] args) throws IOException {
        BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
        while (in.ready()) {
            int res = 0;
            int count = Integer.parseInt(in.readLine());
            if (count == 0) {
                System.out.println(res);
                break;
            }
            String[] strings1 = in.readLine().split(" ");
            int[] nums1 = new int[count * 2];
            for (int i = 0; i < count; i++) {
                nums1[i] = Integer.parseInt(strings1[i]);
            }
            String[] strings2 = in.readLine().split(" ");
            for (int i = 0; i < count; i++) {
                nums1[i + count] = Integer.parseInt(strings2[i]);
            }
            res = merge(nums1, 0,  count - 1, 2 * count - 1);
            System.out.println(res);

        }

        in.close();
    }

    private static int merge(int[] a, int start, int mid, int end) {
        int res = 0;
        int i = start, j = mid + 1;
        while (i <= mid && j <= end) {
            if (a[i] <= a[j])
                i++;
            else {
                j++;
                res += mid - i + 1;
            }
        }
        return res;
    }
}

等差数组

每个元素只能加1或者减1，能否组成，输出修改元素的个数，若不能，则输出-1

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public class Main {
    static int res = -1;

    public static void main(String[] args) throws IOException {
        BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
        in.readLine();
        while (in.ready()) {
            int count = Integer.parseInt(in.readLine());
            String[] strings = in.readLine().split(" ");
            if (count <= 2) {
                System.out.println(0);
                continue;
            }
            int[] nums = new int[count];
            for (int i = 0; i < count; i++) {
                nums[i] = Integer.parseInt(strings[i]);
            }

            if (nums.length < 3) {
                System.out.println("0");
                continue;
            }
            //前两个元素决定等差的差
            int[] cha = {-1, 0, 1};
            for (int i = 0; i < 3; i++) {
                int[] B = nums.clone();
                B[0] += cha[i];
                for (int j = 0; j < 3; j++) {
                    int[] C = B.clone();
                    C[1] += cha[j];
                    dp(C, 2, Math.abs(cha[i]) + Math.abs(cha[j]));
                }
            }
            System.out.println(res);
        }
        in.close();
    }

    //每个元素加1或减1,最少需要修改几个元素
    private static void dp(int[] nums, int index, int change) {
        if (index == nums.length) {
            if (res == -1) {
                res = change;
            } else {
                res = Math.min(change, res);
            }
            return;
        }
        //改变index所在的元素
        int target = 2 * nums[index - 1] - nums[index - 2];
        if (target == nums[index]) {
            dp(nums, index + 1, change);
        } else if (Math.abs(target - nums[index]) <= 1) {
            nums[index] = target;
            dp(nums, index + 1, change + 1);
        }
    }
}