0054. 螺旋矩阵【中等】

• 1. 📝 题目描述
• 2. 🎯 s.1 - 模拟（边界收缩）

1. 📝 题目描述

• leetcode

给你一个 m 行 n 列的矩阵 matrix，请按照顺时针螺旋顺序，返回矩阵中的所有元素。

---

示例 1：

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

输出：[1, 2, 3, 6, 9, 8, 7, 4, 5]

---

示例 2：

输入：matrix = [
  [1, 2, 3, 4],
  [5, 6, 7, 8],
  [9, 10, 11, 12]
]

输出：[1, 2, 3, 4, 8, 12, 11, 10, 9, 5, 6, 7]

---

提示：

• m == matrix.length
• n == matrix[i].length
• 1 <= m, n <= 10
• -100 <= matrix[i][j] <= 100

2. 🎯 s.1 - 模拟（边界收缩）

c

/**
 * Note: The returned array must be malloced, assume caller calls free().
 */
int* spiralOrder(int** matrix, int matrixSize, int* matrixColSize, int* returnSize) {
    int m = matrixSize, n = matrixColSize[0];
    *returnSize = m * n;
    int* result = (int*)malloc(sizeof(int) * m * n);
    int idx = 0;
    int top = 0, bottom = m - 1, left = 0, right = n - 1;

    while (top <= bottom && left <= right) {
        for (int i = left; i <= right; i++) result[idx++] = matrix[top][i];     // 向右
        top++;
        for (int i = top; i <= bottom; i++) result[idx++] = matrix[i][right];   // 向下
        right--;
        if (top <= bottom) {
            for (int i = right; i >= left; i--) result[idx++] = matrix[bottom][i]; // 向左
            bottom--;
        }
        if (left <= right) {
            for (int i = bottom; i >= top; i--) result[idx++] = matrix[i][left];   // 向上
            left++;
        }
    }
    return result;
}

js

/**
 * @param {number[][]} matrix
 * @return {number[]}
 */
var spiralOrder = function (matrix) {
  const result = []
  let top = 0,
    bottom = matrix.length - 1
  let left = 0,
    right = matrix[0].length - 1

  while (top <= bottom && left <= right) {
    for (let i = left; i <= right; i++) result.push(matrix[top][i]) // 向右
    top++
    for (let i = top; i <= bottom; i++) result.push(matrix[i][right]) // 向下
    right--
    if (top <= bottom) {
      for (let i = right; i >= left; i--) result.push(matrix[bottom][i]) // 向左
      bottom--
    }
    if (left <= right) {
      for (let i = bottom; i >= top; i--) result.push(matrix[i][left]) // 向上
      left++
    }
  }
  return result
}

py

class Solution:
    def spiralOrder(self, matrix: List[List[int]]) -> List[int]:
        result = []
        top, bottom = 0, len(matrix) - 1
        left, right = 0, len(matrix[0]) - 1

        while top <= bottom and left <= right:
            for i in range(left, right + 1): result.append(matrix[top][i])      # 向右
            top += 1
            for i in range(top, bottom + 1): result.append(matrix[i][right])    # 向下
            right -= 1
            if top <= bottom:
                for i in range(right, left - 1, -1): result.append(matrix[bottom][i])  # 向左
                bottom -= 1
            if left <= right:
                for i in range(bottom, top - 1, -1): result.append(matrix[i][left])    # 向上
                left += 1
        return result

• 时间复杂度：$O(m\times n)$，每个元素恰好访问一次
• 空间复杂度：$O(1)$，不计输出数组，只使用常数额外空间

算法思路：

• 维护四个边界 top, bottom, left, right，每轮沿「右 -> 下 -> 左 -> 上」顺时针走一圈当前的最外层
• 每遍历完一条边，收缩对应边界（如遍历完顶行后 top++）
• 向左和向上遍历前需额外判断边界是否仍合法，避免单行或单列时重复遍历