【LeetCode & 剑指offer 刷题笔记】目录（持续更新中...）
74. Search a 2D Matrix
Write an efficient algorithm that searches for a value in an
m
x
n
matrix. This matrix has the following properties:
-
Integers in each row are sorted from left to right.
-
The first integer of each row is greater than the last integer of the previous row.
Example 1:
Input:
matrix = [
[1, 3, 5, 7],
[10, 11, 16, 20],
[23, 30, 34, 50]
]
target = 3
Output:
true
Example 2:
Input:
matrix = [
[1, 3, 5, 7],
[10, 11, 16, 20],
[23, 30, 34, 50]
]
target = 13
Output:
false
/*
问题：在有序矩阵中查找某数（每行增序，且行首元素大于前一行的行尾元素，蛇形有序）
方法：将二维数组看成一维数组用二分查找即可
*/
class
Solution
{
public
:
bool
searchMatrix
(
vector
<
vector
<
int
>>&
matrix
,
int
target
)
{
if
(
matrix
.
empty
()
||
matrix
[
0
].
empty
())
return
false
;
int
rows
=
matrix
.
size
();
int
cols
=
matrix
[
0
].
size
();
int
left
=
0
,
right
=
rows
*
cols
-
1
;
while
(
left <= right)
{
int
mid
=
(
left
+
right
)
/
2
;
if
(
matrix
[mid/cols][mid%cols]
<
target
)
left
=
mid
+
1
;
else
if
(
matrix
[
mid
/
cols
][
mid
%
cols
]
>
target
)
right
=
mid
-
1
;
else
return
true
;
}
return
false
;
}
};
240
.
Search a 2D Matrix II
Write an efficient algorithm that searches for a value in an
m
x
n
matrix. This matrix has the following properties:
-
Integers in each row are sorted in ascending from left to right.
-
Integers i n each column are sorted in ascending from top to bottom.
Consider the following matrix:
[
[1, 4, 7, 11, 15],
[2, 5, 8, 12, 19],
[3, 6, 9, 16, 22],
[10, 13, 14, 17, 24],
[18, 21, 23, 26, 30]
]
Example 1:
Input:
matrix, target =
5
Output:
true
Example 2:
Input: matrix, target = 20
Output: false
//问题：有序矩阵中查找(非蛇形有序)
/*
方法一：扫描行，每行进行二分查找
O(r*logc)
*/
class
Solution
{
public
:
bool
searchMatrix
(
vector
<
vector
<
int
>>&
a
,
int
target
)
{
if
(
a
.
empty
()
||
a
[
0
].
empty
())
return
false
;
//若为空，则返回false
int
row
=
a
.
size
();
//行数
int
col
=
a
[
0
].
size
();
//列数
for
(
int
i
=
0
;
i
<
row
;
i
++)
//遍历行，target要处于行首和行尾元素之间
{
if
(
a
[
i
][
0
]<=
target
&&
a
[
i
][
col
-
1
]
>=
target
)
//不能放在for语句里，因为一旦条件不满足就跳出整个for循环了
{
if
(
a
[
i
][
0
]
==
target
||
a
[
i
][
col
-
1
]
==
target
)
return
true
;
//如果行首或行尾元素等于target则返回true
// cout<<"行数为"<<i<<endl;
int
left
=
0
,
right
=
col
-
1
;
//遍历列，用二分查找算法查找
while
(
left
<=
right
)
{
int
mid
=
(
left
+
right
)/
2
;
if
(
a
[
i
][
mid
]
<
target
)
left
=
mid
+
1
;
else
if
(
a
[
i
][
mid
]
>
target
)
right
=
mid
-
1
;
else
return
true
;
}
}
}
return
false
;
}
};
/*
* 方法二：缩小范围，从左下角元素（该行中最小数，该列中最大数）开始比较（利用L型有序）
* 过程：选取矩阵左下角元素，如果等于要查找的数，查找结束，
如果小于目标数，找下一列，j++
如果大于目标数，找上一行，i--
* 此方法效率较第一种高
O(r)或O(c)
*/
class
Solution
{
public
:
bool
searchMatrix
(
vector
<
vector
<
int
>>&
a
,
int
target
)
{
if
(
a
.
empty
()
||
a
[
0
].
empty
())
return
false
;
//若为空，则返回false
int
row
=
a
.
size
();
//行数
int
col
=
a
[
0
].
size
();
//列数
int
i
=
row
-
1
,
j
=
0
;
//从左下角元素开始比较
while
(
i
>=
0
&&
j
<=
col
-
1
)
{
if
(
a
[
i
][
j
]
<
target
)
//如果小于，则找下一列
j
++;
else
if
(
a
[
i
][
j
]
>
target
)
//如果大于，找上一行
i
--;
else
return
true
;
}
return
false
;
}
};