59. 螺旋矩阵 II

题意：

解题思路：

思路：O(n2)。
1. 定义四个方向：上右下左；
2. 左上角开始遍历，走到右边最大值，改变方向往下走...，遍历完所有格子后退出；
3. 我们填充时遍历一次矩阵，矩阵元素个数为 n2个，所以总时间复杂度是 O(n2)；

PHP代码实现：

class Solution {
    /**
     * @param Integer $n
     * @return Integer[][]
     */
    function generateMatrix($n) {
        $top = 0;
        $bottom = $n - 1;
        $left = 0; $right = $n - 1;
        $num = 1;
        $matrix = [];
        while ($top <= $bottom && $left <= $right) {
            for ($i = $left; $i <= $right; $i++) {
                $matrix[$top][$i] = $num++;
            }
            $top++;
            for ($i = $top; $i <= $bottom; $i++) {
                $matrix[$i][$right] = $num++;
            }
            $right--;
            for ($i = $right; $i >= $left; $i--) {
                $matrix[$bottom][$i] = $num++;
            }
            $bottom--;
            for ($i = $bottom; $i >= $top; $i--) {
                $matrix[$i][$left] = $num++;
            }
            $left++;
        }
        foreach ($matrix as $k => $v) {
            ksort($v);
            $matrix[$k] = $v;
        }
        return $matrix;
    }
}

GO代码实现：

func generateMatrix(n int) [][]int {
    matrix := [][]int{}
    for i := 0; i < n; i++{
        matrix = append(matrix, make([]int, n))
    }
    size := n * n
    top, bottom := 0, n-1      //左右边界
    left, right := 0, n-1      //上下边界
    idx := 1
    for idx <= size {
        //向右走
        for i := top; i <= bottom; i++ {
            matrix[left][i] = idx
            idx++
        }
        left++ //上边界收缩
        //向下走
        for i := left; i <= right; i++ {
            matrix[i][bottom] = idx
            idx++
        }
        bottom-- //右边界收缩
        for i := bottom;i >= top; i-- {
            matrix[right][i] = idx
            idx++
        }
        right--
        for i := right; i >= left; i-- {
            matrix[i][top] = idx
            idx++
        }
        top++ 
    }
    return matrix
}

```