Spiral Matrix – LeetCode Solution Java , Python 3, Python 2 , C , C++, Best and Optimal Solutions , All you need.
Given an m x n matrix, return all elements of the matrix in spiral order.
Example 1:
Input: matrix = [[1,2,3],[4,5,6],[7,8,9]] Output: [1,2,3,6,9,8,7,4,5]
Example 2:
Input: matrix = [[1,2,3,4],[5,6,7,8],[9,10,11,12]] Output: [1,2,3,4,8,12,11,10,9,5,6,7]
Constraints:
m == matrix.lengthn == matrix[i].length1 <= m, n <= 10-100 <= matrix[i][j] <= 100
C++ Spiral Matrix LeetCode Solution
vector<int> spiralOrder(vector<vector<int>>& matrix) {
vector<vector<int> > dirs{{0, 1}, {1, 0}, {0, -1}, {-1, 0}};
vector<int> res;
int nr = matrix.size(); if (nr == 0) return res;
int nc = matrix[0].size(); if (nc == 0) return res;
vector<int> nSteps{nc, nr-1};
int iDir = 0; // index of direction.
int ir = 0, ic = -1; // initial position
while (nSteps[iDir%2]) {
for (int i = 0; i < nSteps[iDir%2]; ++i) {
ir += dirs[iDir][0]; ic += dirs[iDir][1];
res.push_back(matrix[ir][ic]);
}
nSteps[iDir%2]--;
iDir = (iDir + 1) % 4;
}
return res;
}
Java Spiral Matrix LeetCode Solution
public class Solution {
public List<Integer> spiralOrder(int[][] matrix) {
List<Integer> res = new ArrayList<Integer>();
if(matrix.length == 0 || matrix[0].length == 0) return res;
int top = 0;
int bottom = matrix.length-1;
int left = 0;
int right = matrix[0].length-1;
while(true){
for(int i = left; i <= right; i++) res.add(matrix[top][i]);
top++;
if(left > right || top > bottom) break;
for(int i = top; i <= bottom; i++) res.add(matrix[i][right]);
right--;
if(left > right || top > bottom) break;
for(int i = right; i >= left; i--) res.add(matrix[bottom][i]);
bottom--;
if(left > right || top > bottom) break;
for(int i = bottom; i >= top; i--) res.add(matrix[i][left]);
left++;
if(left > right || top > bottom) break;
}
return res;
}
}
Python 3 Spiral Matrix LeetCode Solution
def spiralOrder(self, matrix):
return matrix and [*matrix.pop(0)] + self.spiralOrder([*zip(*matrix)][::-1])
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