54. 螺旋矩阵

题意：

解题思路：

思路：
1.从左到右,从上到下,从右到左,从下到上依次遍历每一个数
2.时间复杂度：O(n)，每个元素遍历一次

PHP代码实现：

class Solution {
    /**
     * @param Integer[][] $matrix
     * @return Integer[]
     */
    function spiralOrder($matrix) {
        $ret = [];
        if (count($matrix) == 0 || count($matrix[0]) == 0) return $ret;
        $top = 0; $bottom = count($matrix) - 1;
        $left = 0; $right = count($matrix[0]) - 1;
        while (true) {
            //from left to right
            for ($i = $left; $i <= $right; $i++) {
                array_push($ret, $matrix[$top][$i]);
            }
            $top++;
            if ($left > $right || $top > $bottom) break;
            //from top to bottom
            for ($i = $top; $i <= $bottom; $i++) {
                array_push($ret, $matrix[$i][$right]);
            }
            $right--;
            if ($left > $right || $top > $bottom) break;
            //from right to left
            for ($i = $right; $i >= $left; $i--) {
                array_push($ret, $matrix[$bottom][$i]);
            }
            $bottom--;
            if ($left > $right || $top > $bottom) break;
            //from bottom to top
            for ($i = $bottom; $i >= $top; $i--) {
                array_push($ret, $matrix[$i][$left]);
            }
            $left++;
            if ($left > $right || $top > $bottom) break;
        }
        return $ret;
    }
}

GO代码实现：

func spiralOrder(matrix [][]int) []int {
    if len(matrix) == 0 {
        return nil
    }
    step := 0
    size := len(matrix) * len(matrix[0])
    top, bottom, left, right := 0, len(matrix) - 1, 0, len(matrix[0]) - 1
    result := make([]int, size)
    for step < size {
        for i := left; i <= right && step < size; i++ {
            result[step] = matrix[top][i]
            step++
        }
        top++
        for i := top; i <= bottom && step < size; i++ {
            result[step] = matrix[i][right]
            step++
        }
        right--
        for i := right; i >= left && step < size; i-- {
            result[step] = matrix[bottom][i]
            step++
        }
        bottom--
        for i := bottom; i >= top && step < size; i-- {
            result[step] = matrix[i][left]
            step++
        }
        left++
    }
    return result
}