Given a matrix of m x n elements (m rows, n columns), return all elements of the matrix in spiral order.
Example 1:
Input:
[
[ 1, 2, 3 ],
[ 4, 5, 6 ],
[ 7, 8, 9 ]
]
Output: [1,2,3,6,9,8,7,4,5]
Example 2:
Input:
[
[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]
题意:
按螺旋方式遍历矩阵。
Solution1: Simulation the process and implement it.
code
class Solution {
public List<Integer> spiralOrder(int[][] matrix) {
List<Integer> res = new ArrayList<>();
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;
}
}