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-1
ans := make([]int, rows*columns)
index := 0
for 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 - 1
while 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 - 1
return 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;
}
}