[LeetCode] Unique Paths II

Follow up for "Unique Paths":

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.

For example,

There is one obstacle in the middle of a 3x3 grid as illustrated below.

[
[0,0,0],
[0,1,0],
[0,0,0]
]

The total number of unique paths is

2

.

Note: m and n will be at most 100.

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Array Dynamic Programming

思路：在Unique Paths的基础上，加上obstacleGrid[i-1][j-1]==0 时，f[i][j]=0,另外，注意红色部分的特殊处理。

class Solution {
public:

int uniquePathsWithObstacles(vector<vector<int> > &obstacleGrid)
{
int m = obstacleGrid.size();
int n = obstacleGrid[0].size();

vector<int> row(n + 1, 0);
vector<vector<int> > f(m + 1, row);

if(obstacleGrid[0][0] == 1)
return 0;
f[1][1] = 1;

for(int i = 1; i <= m; i ++)
{
for(int j = 1; j <= n; j ++)
{
if(i == 1 && j == 1)
continue;
if(obstacleGrid[i-1][j-1] == 1)
f[i][j] = 0;
else
f[i][j] = f[i-1][j] + f[i][j-1];
}
}

return f[m]
;
}

};

思路二：上门的思路改成滚动数组

class Solution {
public:

int uniquePathsWithObstacles(vector<vector<int> > &obstacleGrid)
{
int m = obstacleGrid.size();
int n = obstacleGrid[0].size();

if(obstacleGrid[0][0] == 1)
return 0;
vector<int> f(n + 1, 0);

f[1] = 1;

for(int i = 1; i <= m; i ++)
{
for(int j = 1; j <= n; j ++)
{
if(i == 1 && j == 1)
continue;
if(obstacleGrid[i-1][j-1] == 1)
f[j] = 0;
else
f[j] = f[j] + f[j-1];
}
}

return f
;
}

};