letcode算法--21.螺旋矩阵 II

给你一个正整数 n ，生成一个包含 1 到 n2 所有元素，且元素按顺时针顺序螺旋排列的 n x n 正方形矩阵 matrix 。

示例 1：

输入：n = 3
输出：[[1,2,3],[8,9,4],[7,6,5]]

示例 2：

输入：n = 1
输出：[[1]]

方法一：边界控制

class Solution {
    public int[][] generateMatrix(int n) {
        int[][] result = new int[n][n];
        if (n < 1)return null;
        int i = 0,j = 0;
        int left = 0, right = n-1,top = 0, down = n-1;
        int num = 1;
        while (num != n*n + 1) {
            for (i = left; i <= right; i++) {
                result[top][i] = num;
                num++;
            }
            top++;
            for (j = top; j <= down; j ++){
                result[j][right] = num;
                num++;
            }
            right--;
            for (i = right; i >= left; i --){
                result[down][i] = num;
                num++;
            }
            down--;
            for (j = down; j >= top;j --){
                result[j][left] = num;
                num++;
            }
            left++;
        }
        return result;
    }
}

方法二：官方题解

class Solution {
    public int[][] generateMatrix(int n) {
        int maxNum = n * n;
        int curNum = 1;
        int[][] matrix = new int[n][n];
        int row = 0, column = 0;
        int[][] directions = {{0, 1}, {1, 0}, {0, -1}, {-1, 0}}; // 右下左上
        int directionIndex = 0;
        while (curNum <= maxNum) {
            matrix[row][column] = curNum;
            curNum++;
            int nextRow = row + directions[directionIndex][0], nextColumn = column + directions[directionIndex][1];
            if (nextRow < 0 || nextRow >= n || nextColumn < 0 || nextColumn >= n || matrix[nextRow][nextColumn] != 0) {
                directionIndex = (directionIndex + 1) % 4; // 顺时针旋转至下一个方向
            }
            row = row + directions[directionIndex][0];
            column = column + directions[directionIndex][1];
        }
        return matrix;
    }
}