螺旋矩阵

题目

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.

提示:

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

示例

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]

题解

JavaScript

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

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

    for (let i = top; i <= bottom; i++) res.push(matrix[i][right])
    right--

    if (top > bottom || left > right) break

    for (let i = right; i >= left; i--) res.push(matrix[bottom][i])
    bottom--

    for (let i = bottom; i >= top; i--) res.push(matrix[i][left])
    left++
  }

  return res
}

Rust

pub fn spiral_order(matrix: Vec<Vec<i32>>) -> Vec<i32> {
    let (mut top, mut right, mut bottom, mut left) = (0, matrix[0].len() - 1, matrix.len() - 1, 0);
    let mut res = vec![];

    while left <= right && top <= bottom {
        for i in left..=right {
            res.push(matrix[top][i]);
        }
        top += 1;

        for i in top..=bottom {
            res.push(matrix[i][right]);
        }

        // Rust 要时刻小心 usize 类型被减到负值
        if right == 0 {
            break;
        }
        right -= 1;

        if left > right || top > bottom {
            break;
        }

        for i in (left..=right).rev() {
            res.push(matrix[bottom][i]);
        }

        // Rust 要时刻小心 usize 类型被减到负值
        if bottom == 0 {
            break;
        }
        bottom -= 1;

        for i in (top..=bottom).rev() {
            res.push(matrix[i][left]);
        }
        left += 1;
    }

    res

}

Tips

usize 作为索引的数据类型, 经常需要自减操作, 一定要注意不能将 usize 减到负值, 否则它会自动变成 usize::MAX, 这个数必然会造成你的数组越界.

if usize_val > 0 {
    usize_val -= 1;
}