[LeetCode] 54. Spiral Matrix 螺旋矩阵
Given a matrix of m x n elements (m rows, n columns), return all elements of the matrix in spiral order.
For example,
Given the following matrix:
[ [ 1, 2, 3 ], [ 4, 5, 6 ], [ 7, 8, 9 ] ]
You should return [1,2,3,6,9,8,7,4,5].
将一个矩阵按螺旋顺序输出。
解法：用变量left, right, top, bottom记录左，右，顶，底。然后按照左到右，顶到底，右到左，底到顶的顺序循环，把遍历的元素加入到结果。
Java:
public class Solution {
public List<Integer> spiralOrder(int[][] matrix) {
List<Integer> res = new ArrayList<>();
if(matrix == null || matrix.length == 0)
return res;
int rowNum = matrix.length, colNum = matrix[0].length;
int left = 0, right = colNum - 1, top = 0, bot = rowNum - 1;
while(res.size() < rowNum * colNum) {
for(int col = left; col <= right; col++)
res.add(matrix[top][col]);
top++;
if(res.size() < rowNum * colNum) {
for(int row = top; row <= bot; row++)
res.add(matrix[row][right]);
right--;
}
if(res.size() < rowNum * colNum) {
for(int col = right; col >= left; col--)
res.add(matrix[bot][col]);
bot--;
}
if(res.size() < rowNum * colNum) {
for(int row = bot; row >= top; row--)
res.add(matrix[row][left]);
left++;
}
}
return res;
}
}
Python:
class Solution:
# @param matrix, a list of lists of integers
# @return a list of integers
def spiralOrder(self, matrix):
result = []
if matrix == []:
return result
left, right, top, bottom = 0, len(matrix[0]) - 1, 0, len(matrix) - 1
while left <= right and top <= bottom:
for j in xrange(left, right + 1):
result.append(matrix[top][j])
for i in xrange(top + 1, bottom):
result.append(matrix[i][right])
for j in reversed(xrange(left, right + 1)):
if top < bottom:
result.append(matrix[bottom][j])
for i in reversed(xrange(top + 1, bottom)):
if left < right:
result.append(matrix[i][left])
left, right, top, bottom = left + 1, right - 1, top + 1, bottom - 1
return result
C++：
class Solution {
public:
vector<int> spiralOrder(vector<vector<int> > &matrix) {
vector<int> res;
if (matrix.empty() || matrix[0].empty()) return res;
int m = matrix.size(), n = matrix[0].size();
int c = m > n ? (n + 1) / 2 : (m + 1) / 2;
int p = m, q = n;
for (int i = 0; i < c; ++i, p -= 2, q -= 2) {
for (int col = i; col < i + q; ++col)
res.push_back(matrix[i][col]);
for (int row = i + 1; row < i + p; ++row)
res.push_back(matrix[row][i + q - 1]);
if (p == 1 || q == 1) break;
for (int col = i + q - 2; col >= i; --col)
res.push_back(matrix[i + p - 1][col]);
for (int row = i + p - 2; row > i; --row)
res.push_back(matrix[row][i]);
}
return res;
}
};
