Unique Paths II

题目描述：

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.
解题思路：
使用动态规划,dp[i][j]表示从起点（0，0）到（i,j）的不同路径的数目，由于只能向右和向下移动，

所以状态转移方程为：

如果obstacleGrid[i][j]=0,那么dp[i][j]=dp[i-1][j]+dp[i][j-1]，否则dp[i][j]=0;

最后的dp[m-1][n-1]即为所求。

AC代码如下：

class Solution {
public:
int uniquePathsWithObstacles(vector<vector<int>>& obstacleGrid) {
if (obstacleGrid.size() == 0 || obstacleGrid[0].size() == 0) return 0;
int m = obstacleGrid.size();
int n = obstacleGrid[0].size();
vector<vector<int>> dp(m, vector<int>(n, 1));
//initial first row
for (int i = 0; i < n; ++i){
if (obstacleGrid[0][i] == 1) dp[0][i] = 0;
else{
if (i == 0) dp[0][i] = 1;
else dp[0][i] = dp[0][i - 1];
}
}
//initial first col
for (int i = 0; i < m; ++i){
if (obstacleGrid[i][0] == 1) dp[i][0] = 0;
else{
if (i == 0) dp[i][0] = 1;
else dp[i][0] = dp[i - 1][0];
}
}

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