SpiralMatrix

54.problem

Given a matrix of m x n elements (m rows, n columns), return all elements of the matrix in spiral order.

Example 1:

1
2
3
4
5
6
7
8
> Input:
> [
> [ 1, 2, 3 ],
> [ 4, 5, 6 ],
> [ 7, 8, 9 ]
> ]
> Output: [1,2,3,6,9,8,7,4,5]
>

Example 2:

1
2
3
4
5
6
7
8
> 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]
>

key

很简单的循环输出的例子,从【0,0】的位置顺时针扫一圈,然后缩小一圈,继续扫描,不过有一个细节就是第三次第四循环前,要判断一下,防止最后一层循环只有一行

solution

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
public List<Integer> spiralOrder(int[][] matrix) {

List<Integer> ans = new ArrayList<Integer>();
if (matrix.length == 0)
return ans;
int rs = 0, re = matrix.length - 1;// rowStart rowEnd
int cs = 0, ce = matrix[0].length - 1;// columnStart columnEnd
while (rs <= re && cs <= ce) {
for (int i = cs; i <= ce; i++) {
ans.add(matrix[rs][i]);
}
for(int j=rs+1;j<=re;j++) {
ans.add(matrix[j][ce]);
}
if(rs<re&&cs<ce) {
for(int k=ce-1;k>cs;k--) {
ans.add(matrix[re][k]);
}
for(int l=re;l>rs;l--) {
ans.add(matrix[l][cs]);
}
}

rs++;
re--;
cs++;
ce--;
}
return ans;
}

perfect

1
yehh,I'm the perfect