Spiral Matrix I & II
Spiral Matrix I
Given an integer n, generate a square matrix filled with elements from 1 to n^2 in spiral order.
Given n = 3,
You should return the following matrix:
[
[ 1, 2, 3 ],
[ 8, 9, 4 ],
[ 7, 6, 5 ]
]
分析:
从上,右,下,左打印。
public class Solution {
public int[][] generateMatrix(int n) {
int[][] arr = new int[n][n];
int a = ;
int b = n - ;
int k = ;
while (a < b) {
for (int i = a; i <= b; i++) {
arr[a][i] = k++;
}
for (int i = a + ; i <= b - ; i++) {
arr[i][b] = k++;
}
for (int i = b ; i >= a; i--) {
arr[b][i] = k++;
}
for (int i = b - ; i >= a + ; i--) {
arr[i][a] = k++;
}
a++;
b--;
}
// if n is odd, it will be executed, if it is even, it won't be executed.
if (a == b) {
arr[a][b] = k;
}
return arr;
}
}
Spiral Matrix II
Given a matrix of m x n elements (m rows, n columns), return all elements of the matrix in spiral order.
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].
分析:
拿到左上角和右下角的坐标,然后从上,右,下,左打印。然后更新坐标。
public class Solution {
public List<Integer> spiralOrder(int[][] matrix) {
List<Integer> list = new ArrayList<>();
if (matrix == null || matrix.length == || matrix[].length == ) return list;
int a = , b = ;
int x = matrix.length - , y = matrix[].length - ;
while (a <= x && b <= y) {
// top row
for (int i = b; i <= y; i++) {
list.add(matrix[a][i]);
}
// right column
for (int i = a + ; i <= x - ; i++) {
list.add(matrix[i][y]);
}
// bottom row
if (a != x) {
for (int i = y; i >= b; i--) {
list.add(matrix[x][i]);
}
}
// left column
if (b != y) {
for (int i = x - ; i >= a + ; i--) {
list.add(matrix[i][b]);
}
}
a++;
b++;
x--;
y--;
}
return list;
}
}
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