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问题描述

小C发现了一种奇特的图案，叫做螺旋阵列。它由一串0和1组成，看起来像一个由外向内旋转的图形。小C想知道，能否根据给定的宽度来生成这样一个螺旋图案。

例如，宽度为5时的螺旋阵列如下：

11111
00001
11101
10001
11111

宽度为10时的螺旋阵列如下：

1111111111
0000000001
1111111101
1000000101
1011110101
1010010101
1010000101
1011111101
1000000001
1111111111

小C想知道，对于任意给定的宽度 n，是否能生成对应的螺旋图案，并且以一个二维数组的形式输出。

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测试样例

样例1：

输入：width = 5
输出：[[1, 1, 1, 1, 1], [0, 0, 0, 0, 1], [1, 1, 1, 0, 1], [1, 0, 0, 0, 1], [1, 1, 1, 1, 1]]

样例2：

输入：width = 8
输出：[[1, 1, 1, 1, 1, 1, 1, 1], [0, 0, 0, 0, 0, 0, 0, 1], [1, 1, 1, 1, 1, 1, 0, 1], [1, 0, 0, 0, 0, 1, 0, 1], [1, 0, 1, 0, 0, 1, 0, 1], [1, 0, 1, 1, 1, 1, 0, 1], [1, 0, 0, 0, 0, 0, 0, 1], [1, 1, 1, 1, 1, 1, 1, 1]]

样例3：

输入：width = 2
输出：[[1, 1], [0, 1]]

 

#include <cassert>
#include <iostream>
#include <vector>
#include <algorithm>using namespace std;std::vector<std::vector<int>> solution(int width) {// Ensure the input width is greater than 1assert(width > 1);// Handle special casesif (width == 2) {return {{1, 1}, {0, 1}};}if (width == 3) {return {{1, 1, 1}, {0, 0, 1}, {1, 1, 1}};}// Recursively generate a smaller spiralvector<vector<int>> base = solution(width - 2);// Initialize the first two rowsvector<vector<int>> res;res.push_back(vector<int>(width, 1));res.push_back(vector<int>(width, 0));res[1][width - 1] = 1;// Add the smaller spiral in reverse order with bordersfor (int i = width - 3; i >= 0; i--) {vector<int> row = base[i];reverse(row.begin(), row.end());row.push_back(0);row.push_back(1);res.push_back(row);}// Fix the second last element of the last rowres[res.size() - 1][res[0].size() - 2] = 1;return res;
}int main() {// You can add more test cases herestd::vector<std::vector<int>> expected1 = {{1, 1, 1, 1, 1},{0, 0, 0, 0, 1},{1, 1, 1, 0, 1},{1, 0, 0, 0, 1},{1, 1, 1, 1, 1}};std::vector<std::vector<int>> expected2 = {{1, 1, 1, 1, 1, 1, 1, 1}, {0, 0, 0, 0, 0, 0, 0, 1},{1, 1, 1, 1, 1, 1, 0, 1}, {1, 0, 0, 0, 0, 1, 0, 1},{1, 0, 1, 0, 0, 1, 0, 1}, {1, 0, 1, 1, 1, 1, 0, 1},{1, 0, 0, 0, 0, 0, 0, 1}, {1, 1, 1, 1, 1, 1, 1, 1}};std::vector<std::vector<int>> expected3 = {{1, 1}, {0, 1}};std::cout << (solution(5) == expected1) << std::endl;std::cout << (solution(8) == expected2) << std::endl;std::cout << (solution(2) == expected3) << std::endl;return 0;
}