【LeetCode & 剑指offer 刷题笔记】目录（持续更新中...）
Unique Paths（系列）
Unique Paths
A robot is located at the top-left corner of a
m
x
n
grid (marked 'Start' in the diagram below).
The robot can
only move either down or right a
t any point in time. The robot is trying to reach the bottom-right corner of the grid (marked 'Finish' in the diagram below).
How many possible unique paths are there?
Above is a 7 x 3 grid. How many possible unique paths are there?
Note:
m
and
n
will be at most 100.
Example 1:
Input:
m = 3, n = 2
Output:
3
Explanation:
From the top-left corner, there are a total of 3 ways to reach the bottom-right corner:
1. Right -> Right -> Down
2. Right -> Down -> Right
3. Down -> Right -> Right
Example 2:
Input:
m = 7, n = 3
Output:
28
//问题：街区路径问题(与LCS最长公共子序列问题类似) 求机器人从起点到终点的路径数量
//方法一：设P[i][j] 为到坐标（i,j）的路径总数，则由于机器人
只能向右或者向下
，故有P[i][j] = P[i - 1][j] + P[i][j - 1]
//O(n^2) O(m*n)
class
Solution
{
public
:
int
uniquePaths
(
int
m
,
int
n
)
{
vector
<
vector
<
int
>>
path
(
m
,
vector
<
int
>(
n
,
1
));
//(1,0)和（0,1）处初始化为1
for
(
int
i
=
1
;
i
<
m
;
i
++)
//i从1开始，扫描行
for
(
int
j
=
1
;
j
<
n
;
j
++)
//扫描列
path
[i][j] = path[i-1][j] + path[i][j-1];
return
path
[
m
-
1
][
n
-
1
];
}
};
/*
* 方法一改进
用一个列向量存当前需要的元素即可，将空间复杂度优化至O（min(m,n)）
* 例子
1 1 1 1 1 1
1 2 3 4 5 6
1 3 6 10 15 21
dp[i][j] = dp[i - 1][j] + dp[i][j - 1]
dp[i][j],dp[i-1][j] ---> dp[j]
dp[i][j-1] ---> dp[j-1]
*/
class Solution
{
public
:
int
uniquePaths
(
int
m
,
int
n
)
{
vector
<
int
>
dp
(
n
,
1
);
//进一步改进可以选择min(m,n)
for
(
int
i
=
1
;
i
<
m
;
++
i
)
//扫描行
{
for
(
int
j
=
1
;
j
<
n
;
++
j
)
//扫描列
{
dp
[
j
]
+=
dp
[
j
-
1
];
}
}
return
dp
[
n
-
1
];
}
};
63
.
Unique Paths II
Now consider if some obstacles are added to the grids. How many unique paths would there be?
An obstacle and empty space is marked as
1
and
0
respectively in the grid.
Note:
m
and
n
will be at most 100.
Example 1:
Input:
[
[0,0,0],
[0,1,0],
[0,0,0]
]
Output:
2
Explanation:
There is one obstacle in the middle of the 3x3 grid above.
There are two ways to reach the bottom-right corner:
1. Right -> Right -> Down -> Down
2. Down -> Down -> Right -> Right
/*
问题：有障碍物的栅格中，机器人从起始点到终点的路径数量
方法：动态规划
对于有障碍物的栅格中，dp值置0
*/
class
Solution
{
public
:
int
uniquePathsWithObstacles
(
vector
<
vector
<
int
>>&
obstacleGrid
)
{
if
(
obstacleGrid
.
empty
()
||
obstacleGrid
[
0
].
empty
()
||
obstacleGrid
[
0
][
0
]
==
1
)
return
0
;
//矩阵为空或者起始点为障碍点时，退出
int
m
=
obstacleGrid
.
size
();
int
n
=
obstacleGrid
[
0
].
size
();
vector
<
vector
<
int
>>
dp
(
m
,
vector
<
int
>(
n
));
//初始化为0
for
(
int
i
=
0
;
i
<
m
;
i
++)
//从0开始扫描，因为任何点都有可能为障碍点(如（0,1）和（1,0）)
{
for
(
int
j
=
0
;
j
<
n
;
j
++)
{
if
(
obstacleGrid
[
i
][
j
]
==
1
)
dp
[
i
][
j
]
=
0
;
//有障碍的地方dp值置0
else
if
(
i
==
0
&&
j
==
0
)
dp
[
i
][
j
]
=
1
;
//起始点到起始点路径数为1（若无障碍）
else
if
(
i
==
0
&&
j
>
0
)
dp
[
i
][
j
]
=
dp
[
i
][
j
-
1
];
//第一行的处理
else
if
(
i
>
0
&&
j
==
0
)
dp
[
i
][
j
]
=
dp
[
i
-
1
][
j
];
//第一列的处理
else
dp
[
i
][
j
]
=
dp
[
i
-
1
][
j
]
+
dp
[
i
][
j
-
1
];
//其他情况的处理
}
}
return
dp
.
back
().
back
();
//返回末尾点
}
};