题目描述

题目：矩阵的螺旋遍历

给定一个m x n的矩阵（m行，n列），请按照螺旋顺序返回矩阵中的所有元素。螺旋顺序指的是从左上角开始，首先向右遍历，然后向下，接着向左，最后向上，以此类推，直到遍历完所有元素。

示例：

输入矩阵：

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

输出：[1, 2, 3, 6, 9, 8, 7, 4, 5]

PHP 示例代码

function spiralOrder($matrix) {
    if (empty($matrix)) return [];

    $result = [];
    $top = 0;
    $bottom = count($matrix) - 1;
    $left = 0;
    $right = count($matrix[0]) - 1;

    while ($left <= $right && $top <= $bottom) {
        // Traverse right
        for ($i = $left; $i <= $right; $i++) {
            $result[] = $matrix[$top][$i];
        }
        $top++;

        // Traverse down
        for ($i = $top; $i <= $bottom; $i++) {
            $result[] = $matrix[$i][$right];
        }
        $right--;

        if ($top <= $bottom) {
            // Traverse left
            for ($i = $right; $i >= $left; $i--) {
                $result[] = $matrix[$bottom][$i];
            }
            $bottom--;
        }

        if ($left <= $right) {
            // Traverse up
            for ($i = $bottom; $i >= $top; $i--) {
                $result[] = $matrix[$i][$left];
            }
            $left++;
        }
    }

    return $result;
}

// 示例用法
$matrix = [
    [1, 2, 3],
    [4, 5, 6],
    [7, 8, 9]
];
print_r(spiralOrder($matrix));

Python 示例代码

def spiralOrder(matrix):
    if not matrix:
        return []

    result = []
    top, bottom = 0, len(matrix) - 1
    left, right = 0, len(matrix[0]) - 1

    while left <= right and top <= bottom:
        # Traverse right
        for i in range(left, right + 1):
            result.append(matrix[top][i])
        top += 1

        # Traverse down
        for i in range(top, bottom + 1):
            result.append(matrix[i][right])
        right -= 1

        if top <= bottom:
            # Traverse left
            for i in range(right, left - 1, -1):
                result.append(matrix[bottom][i])
            bottom -= 1

        if left <= right:
            # Traverse up
            for i in range(bottom, top - 1, -1):
                result.append(matrix[i][left])
            left += 1

    return result

# 示例用法
matrix = [
    [1, 2, 3],
    [4, 5, 6],
    [7, 8, 9]
]
print(spiralOrder(matrix))

JavaScript 示例代码

function spiralOrder(matrix) {
    if (!matrix.length) return [];

    let result = [];
    let top = 0, bottom = matrix.length - 1;
    let left = 0, right = matrix[0].length - 1;

    while (left <= right && top <= bottom) {
        // Traverse right
        for (let i = left; i <= right; i++) {
            result.push(matrix[top][i]);
        }
        top++;

        // Traverse down
        for (let i = top; i <= bottom; i++) {
            result.push(matrix[i][right]);
        }
        right--;

        if (top <= bottom) {
            // Traverse left
            for (let i = right; i >= left; i--) {
                result.push(matrix[bottom][i]);
            }
            bottom--;
        }

        if (left <= right) {
            // Traverse up
            for (let i = bottom; i >= top; i--) {
                result.push(matrix[i][left]);
            }
            left++;
        }
    }

    return result;
}

// 示例用法
const matrix = [
    [1, 2, 3],
    [4, 5, 6],
    [7, 8, 9]
];
console.log(spiralOrder(matrix));

这些示例代码均实现了矩阵的螺旋遍历，并给出了具体的示例用法。在面试中，你可以根据面试官的要求选择使用哪种编程语言进行解答。