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.

题意：如果地图上有障碍物，此障碍物不能通过 在数组中用1表示障碍物，求从起点到终点有多少种走法。

思路
   1. 如果没有障碍

val[i][0] = 1
val[0][j] = 1
val[i][j] = val[i-1][j] + val[i][j-1]

2. 有了障碍后
如果obstacle[i][j] = 1
val[i][j] = 1
否则
tmp = obstacle[i-1][j] == 1 ? 0 : val[i-1][j]
tmp = obstacle[i][j-1] == 1 ? tmp : tmp + val[i-1][j-1]
　　　val[i][j] = tmp

class Solution {
public:
int uniquePathsWithObstacles(vector<vector<int> > &obstacleGrid) {
int row = obstacleGrid.size();
int col = obstacleGrid[0].size();
int token = 1;
int val[row][col];
for (int j = 0; j < col; ++j)
{
if(obstacleGrid[0][j] == 1)
token = 0;
val[0][j] = token;

}
token = 1;
for (int i = 0; i < row; ++i)
{
if(obstacleGrid[i][0] == 1)
token = 0;
val[i][0] = token;
}
for (int i = 1; i < row; ++i)
{
for(int j = 1; j < col; ++j)
{
if (obstacleGrid[i][j] == 1)
val[i][j] = 0;
else
{
int tmp = obstacleGrid[i-1][j] == 1 ? 0 :val[i-1][j];
tmp = obstacleGrid[i][j-1] == 1 ? tmp : tmp + val[i][j-1];
val[i][j] = tmp;
}
}
}
return val[row-1][col-1];
}
};