题目
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.
分析
带障碍的路径数计算问题，这个题目与上一题的区别之处在于，m*n的矩阵中有部分设置了障碍，当然有障碍的地方不能通过；
虽然设立了障碍，该题目的本质仍然是一个动态规划问题，我们只需要增加判断当前点是否有障碍的代码即可，若有障碍那么此处不能通行，自然f(i,j)=0，对于其他点，依然用上一题目的推导公式即可！
AC代码
//直接用非递归算法求解
class Solution {
public:
int uniquePathsWithObstacles(vector<vector<int>>& obstacleGrid) {
if (obstacleGrid.empty())
return 0;
int m = obstacleGrid.size();
int n = obstacleGrid[0].size();
vector<vector<int> > ret(m, vector<int>(n, 0));
//矩阵首列
for (int i = 0; i < m; i++)
{
//无障碍，则有一条路径，否则不通
if (obstacleGrid[i][0] != 1)
ret[i][0] = 1;
else
break;
}//for
//矩阵首行
for (int j = 0; j < n; j++)
{
//无障碍，则有一条路径，否则不通
if (obstacleGrid[0][j] != 1)
ret[0][j] = 1;
else
break;
}//for
//其余位置
for (int i = 1; i < m; i++)
{
for (int j = 1; j < n; j++)
{
//当前位置为障碍，则到此处路径数为0
if (obstacleGrid[i][j] == 1)
ret[i][j] = 0;
else{
ret[i][j] = ret[i][j - 1] + ret[i - 1][j];
}//else
}//for
}//for
return ret[m - 1][n - 1];
}//uniques
};