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算法leetcode｜54. 螺旋矩阵（rust重拳出击）

news 2025/8/12 4:27:04

文章目录

54. 螺旋矩阵：
- 样例 1：
- 样例 2：
- 提示：
分析：
题解：
- rust：
- go：
- c++：
- python：
- java：
- - 每次循环移动一步：
  - 每次循环完成一个顺时针：

54. 螺旋矩阵：

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。

样例 1：

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

样例 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

分析：

面对这道算法题目，二当家的陷入了沉思。
可以每次循环移动一步，判断移到边界就变换方向。
也可以每次循环都换完4次方向，也就是完成一次顺时针，然后缩圈。

题解：

rust：

impl Solution {pub fn spiral_order(matrix: Vec<Vec<i32>>) -> Vec<i32> {let mut ans = Vec::new();let (rows, columns) = (matrix.len(), matrix[0].len());let (mut left, mut right, mut top, mut bottom) = (0, columns - 1, 0, rows - 1);while left <= right && top <= bottom {(left..right + 1).for_each(|column| {ans.push(matrix[top][column]);});(top + 1..bottom + 1).for_each(|row| {ans.push(matrix[row][right]);});if left < right && top < bottom {(left..right).rev().for_each(|column| {ans.push(matrix[bottom][column]);});(top + 1..bottom).rev().for_each(|row| {ans.push(matrix[row][left]);});}if right == 0 || bottom == 0 {break;}left += 1;right -= 1;top += 1;bottom -= 1;}return ans;}
}

go：

func spiralOrder(matrix [][]int) []int {rows, columns := len(matrix), len(matrix[0])left, right, top, bottom := 0, columns-1, 0, rows-1ans := make([]int, rows*columns)index := 0for left <= right && top <= bottom {for column := left; column <= right; column++ {ans[index] = matrix[top][column]index++}for row := top + 1; row <= bottom; row++ {ans[index] = matrix[row][right]index++}if left < right && top < bottom {for column := right - 1; column >= left; column-- {ans[index] = matrix[bottom][column]index++}for row := bottom - 1; row > top; row-- {ans[index] = matrix[row][left]index++}}left++right--top++bottom--}return ans
}

c++：

class Solution {
public:vector<int> spiralOrder(vector<vector<int>>& matrix) {vector<int> ans;int rows = matrix.size(), columns = matrix[0].size();int left = 0, right = columns - 1, top = 0, bottom = rows - 1;while (left <= right && top <= bottom) {for (int column = left; column <= right; ++column) {ans.emplace_back(matrix[top][column]);}for (int row = top + 1; row <= bottom; ++row) {ans.emplace_back(matrix[row][right]);}if (left < right && top < bottom) {for (int column = right - 1; column >= left; --column) {ans.emplace_back(matrix[bottom][column]);}for (int row = bottom - 1; row > top; --row) {ans.emplace_back(matrix[row][left]);}}++left;--right;++top;--bottom;}return ans;}
};

python：

class Solution:def spiralOrder(self, matrix: List[List[int]]) -> List[int]:ans = list()rows, columns = len(matrix), len(matrix[0])left, right, top, bottom = 0, columns - 1, 0, rows - 1while left <= right and top <= bottom:for column in range(left, right + 1):ans.append(matrix[top][column])for row in range(top + 1, bottom + 1):ans.append(matrix[row][right])if left < right and top < bottom:for column in range(right - 1, left - 1, -1):ans.append(matrix[bottom][column])for row in range(bottom - 1, top, -1):ans.append(matrix[row][left])left, right, top, bottom = left + 1, right - 1, top + 1, bottom - 1return ans

java：

每次循环移动一步：

class Solution {public List<Integer> spiralOrder(int[][] matrix) {List<Integer> ans = new ArrayList<Integer>();final int     rows             = matrix.length, columns = matrix[0].length;int           left             = 0, right = columns - 1, top = 0, bottom = rows - 1;final int     total            = rows * columns;int           row              = 0, column = 0;final int[][] moveDirections   = {{0, 1}, {1, 0}, {0, -1}, {-1, 0}};final int[][] borderDirections = {{1, 0, 0, 0}, {0, 0, 1, 0}, {0, -1, 0, 0}, {0, 0, 0, -1}};int           directionIndex   = 0;for (int i = 0; i < total; i++) {ans.add(matrix[row][column]);int nextRow = row + moveDirections[directionIndex][0], nextColumn = column + moveDirections[directionIndex][1];if (nextRow < top || nextRow > bottom || nextColumn < left || nextColumn > right) {// 变换方向directionIndex = (directionIndex + 1) % 4;// 修改边界left += borderDirections[directionIndex][0];right += borderDirections[directionIndex][1];top += borderDirections[directionIndex][2];bottom += borderDirections[directionIndex][3];}row += moveDirections[directionIndex][0];column += moveDirections[directionIndex][1];}return ans;}
}

每次循环完成一个顺时针：

class Solution {public List<Integer> spiralOrder(int[][] matrix) {List<Integer> ans = new ArrayList<Integer>();final int rows = matrix.length, columns = matrix[0].length;int       left = 0, right = columns - 1, top = 0, bottom = rows - 1;while (left <= right && top <= bottom) {for (int column = left; column <= right; ++column) {ans.add(matrix[top][column]);}for (int row = top + 1; row <= bottom; ++row) {ans.add(matrix[row][right]);}if (left < right && top < bottom) {for (int column = right - 1; column >= left; --column) {ans.add(matrix[bottom][column]);}for (int row = bottom - 1; row > top; --row) {ans.add(matrix[row][left]);}}++left;--right;++top;--bottom;}return ans;}
}