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- Spiral Matrix
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].
class Solution {
public List<Integer> spiralOrder(int[][] matrix) {
List<Integer> res = new ArrayList<>();
if (matrix.length == 0)
return res;
int rowBegin = 0, colBegin = 0;
int rowEnd = matrix.length - 1;
int colEnd = matrix[0].length - 1;
while (rowBegin <= rowEnd && colBegin <= colEnd) {
// Traverse right
for (int j = colBegin; j <= colEnd; j ++) {
res.add(matrix[rowBegin][j]);
}
rowBegin ++;
// Traverse down
for (int i = rowBegin; i <= rowEnd; i ++) {
res.add(matrix[i][colEnd]);
}
colEnd --;
if (rowBegin <= rowEnd) {
// Traverse left
for (int j = colEnd; j >= colBegin; j --) {
res.add(matrix[rowEnd][j]);
}
rowEnd --;
}
if (colBegin <= colEnd) {
// Traverse up
for (int i = rowEnd; i >= rowBegin; i --) {
res.add(matrix[i][colBegin]);
}
colBegin ++;
}
}
return res;
}
}
- Spiral Matrix II
Given an integer n, generate a square matrix filled with elements from 1 to n2 in spiral order.
For example,
Given n = 3,
You should return the following matrix:
[
[ 1, 2, 3 ],
[ 8, 9, 4 ],
[ 7, 6, 5 ]
]
class Solution {
public int[][] generateMatrix(int n) {
int[][] matrix = new int[n][n];
if (n == 0) return matrix;
int rowBegin = 0, colBegin = 0;
int rowEnd = n - 1, colEnd = n - 1;
int num = 1;
while (rowBegin <= rowEnd && colBegin <= colEnd) {
// Traverse right
for (int j = colBegin; j <= colEnd; j ++) {
matrix[rowBegin][j] = num ++;
}
rowBegin ++;
// Traverse down
for (int i = rowBegin; i <= rowEnd; i ++) {
matrix[i][colEnd] = num ++;
}
colEnd --;
if (rowBegin <= rowEnd) {
// Traverse left
for (int j = colEnd; j >= colBegin; j --) {
matrix[rowEnd][j] = num ++;
}
rowEnd --;
}
if (colBegin <= colEnd) {
// Traverse up
for (int i = rowEnd; i >= rowBegin; i --) {
matrix[i][colBegin] = num ++;
}
colBegin ++;
}
}
return matrix;
}
}