【LeetCode & 剑指offer 刷题笔记】目录（持续更新中...）
54. Spiral Matrix
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]
//以螺旋顺序输出矩阵元素（顺时针）
/*
确定某一层矩形对角点坐标，(begin,begin)、(rowend,colend)
扫描顺序如下：
-----→
↑ |
| |
| |
←----↓
*/
class
Solution
{
public
:
vector
<
int
>
spiralOrder
(
vector
<
vector
<
int
>>&
a
)
{
vector
<
int
>
result
;
if
(
a
.
empty
())
return
result
;
int
begin
=
0
;
//起始点坐标
int
rowend
=
a
.
size
()
-
1
;
//对角点坐标
int
colend
=
a
[
0
].
size
()
-
1
;
while
(
begin
<=
rowend
&&
begin
<=
colend
)
{
for
(
int
j
=
begin
;
j
<=
colend
;
j
++)
//输出上行 i = begin,j = begin ~ colend
{
result
.
push_back
(
a
[
begin
][
j
]);
}
for
(
int
i
=
begin
+
1
;
i
<=
rowend
;
i
++)
//输出右边列 i = begin+1 ~ rowend, j = colend
{
result
.
push_back
(
a
[
i
][
colend
]);
}
if
(
begin
<
rowend
)
//防止最后只有一行
{
for
(
int
j
=
colend
-
1
;
j
>=
begin
;
j
--)
//输出下行 若begin<rowend, i = rowend, j = colend-1 ~ begin
{
result
.
push_back
(
a
[
rowend
][
j
]);
}
}
if
(
begin
<
colend
)
//防止最后只有一列
{
for
(
int
i
=
rowend
-
1
;
i
>=
begin
+
1
;
i
--)
//输出左边列 若begin<colend, i = rowend-1 ~ begin+1, j = begin
{
result
.
push_back
(
a
[
i
][
begin
]);
}
}
begin
++;
//缩至内圈
rowend
--;
colend
--;
}
return
result
;
}
};
59
.
Spiral Matrix II
Given a positive integer
n
, generate a square matrix filled with elements from 1 to
n
2
in spiral order.
Example:
Input: 3
Output:
[
[ 1, 2, 3 ],
[ 8, 9, 4 ],
[ 7, 6, 5 ]
]
/*
问题：按螺旋的顺序来填数
与
Spiral Matrix
类似
*/
class
Solution
{
public
:
vector
<
vector
<
int
>>
generateMatrix
(
int
n
)
{
vector
<
vector
<
int
>>
a
(
n
,
vector
<
int
>(
n
));
if
(
n
<
1
)
return
a
;
int
begin
=
0
;
//
起始点坐标
int
rowend
=
n
-
1
;
//
对角点坐标
int
colend
=
n
-
1
;
int
val
=
1
;
while
(
begin
<=
rowend
&&
begin
<=
colend
)
{
for
(
int
j
=
begin
;
j
<=
colend
;
j
++)
//
赋值上行
i = begin,j = begin ~ colend
{
a
[
begin
][
j
]
=
val
++;
//++
在后为先用后加，故
val
初值为
1
}
for
(
int
i
=
begin
+
1
;
i
<=
rowend
;
i
++)
//
赋值右边列
i = begin+1 ~ rowend, j = colend
{
a
[
i
][
colend
]
=
val
++;
}
if
(
begin
<
rowend
)
//
防止最后只有一行
{
for
(
int
j
=
colend
-
1
;
j
>=
begin
;
j
--)
//
赋值下行
若
begin<rowend, i = rowend, j = colend-1 ~ begin
{
a
[
rowend
][
j
]
=
val
++;
}
}
if
(
begin
<
colend
)
//
防止最后只有一列
{
for
(
int
i
=
rowend
-
1
;
i
>=
begin
+
1
;
i
--)
//
赋值左边列
若
begin<colend, i = rowend-1 ~ begin+1, j = begin
{
a
[
i
][
begin
]
=
val
++;
}
}
begin
++;
//
缩至内圈
rowend
--;
colend
--;
}
return
a
;
}
};