题目

给定一个正整数 n，生成一个包含 1 到 n^2 所有元素，且元素按顺时针顺序螺旋排列的正方形矩阵。

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

题解

这个题目也比较简单, 和第54题类似:
这个题目很简单,上下左右分别用四个变量去标志：

上：top
下：bottom
左：left
右: right

就按照四步走就可以:

left -> right
top -> bottom
right -> left
bottom -> top

python

class Solution:
    def generateMatrix(self, n):
        """
        :type n: int
        :rtype: List[List[int]]
        """
        matrix = [[0 for j in range(n)] for i in range(n)]
        top = 0
        bottom = n - 1
        left = 0
        right = n - 1
        count = 1
        while (True):
            for i in range(left, right + 1):
                matrix[top][i] = count
                count += 1
            top += 1
            if left > right or top > bottom:
                break
            for i in range(top, bottom + 1):
                matrix[i][right] = count
                count += 1
            right -= 1
            if left > right or top > bottom:
                break
            for j in range(left, right + 1)[::-1]:
                matrix[bottom][j] = count
                count += 1
            bottom -= 1
            if left > right or top > bottom:
                break
            for i in range(top, bottom + 1)[::-1]:
                matrix[i][left] = count
                count += 1
            left += 1
            if left > right or top > bottom:
                break
        return matrix

java

class Solution {
    public int[][] generateMatrix(int n) {
         int[][] matrix = new int[n][n];
        int top = 0;
        int bottom = matrix.length - 1;
        int left = 0;
        int right = matrix[0].length - 1;
        int count = 1;
        while (true) {
            for (int i = left; i <= right; i++) matrix[top][i] = count++;
            top++;
            if (left > right || top > bottom) break;

            for (int i = top; i <= bottom; i++) matrix[i][right] = count++;
            right--;
            if (left > right || top > bottom) break;

            for (int i = right; i >= left; i--) matrix[bottom][i] = count++;
            bottom--;
            if (left > right || top > bottom) break;

            for (int i = bottom; i >= top; i--) matrix[i][left] = count++;
            left++;
            if (left > right || top > bottom) break;
        }

        return matrix;
    }
}

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