Spiral Matrix II

Given an integer n, generate a square matrix filled with elements from 1 ton² in spiral order.

For example,
Given n = 3,

You should return the following matrix:

[ [ 1, 2, 3 ], [ 8, 9, 4 ], [ 7, 6, 5 ]]

1. Works only for even n. 因为无法处理最后的中心数

public int[][] generateMatrix(int n) {// Start typing your Java solution below// DO NOT write main() functionint[][] res = new int[n][n];int max = n * n;int counter = 1;int layer = 0;while(true){for(int i = layer; i < n - layer - 1; i++){res[layer][i] = counter;counter++;}if(counter > max)break;for(int i = layer; i < n - layer - 1; i++){res[i][n - layer - 1] = counter;counter++;}if(counter > max)break;for(int i = n - layer - 1; i >= layer + 1; i--){res[n - layer - 1][i] = counter;counter++;}if(counter > max)break;for(int i = n - layer - 1; i >= layer + 1; i--){res[i][layer] = counter;counter++;}if(counter > max)break;layer++;}return res;}

2. 稍微处理下边角，通过。

public class Solution {    public int[][] generateMatrix(int n) {// Start typing your Java solution below// DO NOT write main() functionint[][] res = new int[n][n];int max = n * n;int counter = 1;int layer = 0;while(true){for(int i = layer; i <= n - layer - 1; i++){res[layer][i] = counter;counter++;}if(counter > max)break;for(int i = layer + 1; i < n - layer - 1; i++){res[i][n - layer - 1] = counter;counter++;}if(counter > max)break;for(int i = n - layer - 1; i >= layer; i--){res[n - layer - 1][i] = counter;counter++;}if(counter > max)break;for(int i = n - layer - 2; i >= layer + 1; i--){res[i][layer] = counter;counter++;}if(counter > max)break;layer++;}return res;}}

3. Better coding style

public class Solution {    public int[][] generateMatrix(int n) {// Start typing your Java solution below// DO NOT write main() functionint[][] res = new int[n][n];if (n == 0)return res;int num = 0;int layer = n / 2;for (int i = 0; i < layer; ++i) {int j;for (j = i; j < n - 1 - i; ++j)res[i][j] = ++num;for (j = i; j < n - 1 - i; ++j)res[j][n - 1 - i] = ++num;for (j = n - 1 - i; j > i; --j)res[n - 1 - i][j] = ++num;for (j = n - 1 - i; j > i; --j)res[j][i] = ++num;}if (n % 2 == 1)res[n / 2][n / 2] = ++num;return res;}}