题目要求：
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.
就是在一个矩阵中，求从左上到右下单元的路径个数，每次只能向下或向右移动。0表示可以通过，1表示障碍。
思路：
本题依然是1个DP问题。设dp[i,j]表示从左上到矩阵的[i,j]位置的路径个数。
1. 当行或列只有1时，如果有障碍，则返回0，否则返回1。
当row > 1 且 col > 1:
2.1 分别初始化dp的第一行(i)和第一列(j)，即dp[i,0]和dp[0,j]：如果发现障碍，即i==1或j==1，把则dp[i...n,0]和dp[0,j...n]设为0，没有发现障碍时初始化为1。
2.2 如果obstacleGrid[i,j]为1，即当前为障碍，则dp[i,j]为0，否则：
dp[i, j] = dp[i-1, j] + dp[i, j-1];
实现代码：
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.
就是在一个矩阵中，求从左上到右下单元的路径个数，每次只能向下或向右移动。0表示可以通过，1表示障碍。
思路：
本题依然是1个DP问题。设dp[i,j]表示从左上到矩阵的[i,j]位置的路径个数。
1. 当行或列只有1时，如果有障碍，则返回0，否则返回1。
当row > 1 且 col > 1:
2.1 分别初始化dp的第一行(i)和第一列(j)，即dp[i,0]和dp[0,j]：如果发现障碍，即i==1或j==1，把则dp[i...n,0]和dp[0,j...n]设为0，没有发现障碍时初始化为1。
2.2 如果obstacleGrid[i,j]为1，即当前为障碍，则dp[i,j]为0，否则：
dp[i, j] = dp[i-1, j] + dp[i, j-1];
实现代码：
public class Solution {
public int UniquePathsWithObstacles(int[,] obstacleGrid) {
var row = obstacleGrid.GetLength(0);
var col = obstacleGrid.GetLength(1);
if(row < 1 || obstacleGrid[0,0] == 1){
return 0;
}
if(row == 1){
for(var i = 0; i < col; i ++){
if(obstacleGrid[0,i] == 1){
return 0;
}
}
return 1;
}
if(col == 1){
for(var i = 0; i < row; i ++){
if(obstacleGrid[i,0] == 1){
return 0;
}
}
return 1;
}
var dp = new int[row, col];
for(var i = 0;i < col; i++){
if(obstacleGrid[0,i] == 1){
while(i < col){
dp[0,i] = 0;
i++;
}
break;
}
else{
dp[0,i] = 1;
}
}
for(var i = 0;i < row; i++){
if(obstacleGrid[i,0] == 1){
while(i < row){
dp[i,0] = 0;
i++;
}
break;
}
else{
dp[i,0] = 1;
}
}
for(var i = 1;i < row; i++){
for(var j = 1;j < col; j++){
if(obstacleGrid[i,j] == 1){
dp[i,j] = 0;
}
else{
dp[i, j] = dp[i-1, j] + dp[i, j-1];
}
}
}
return dp[row-1, col-1];
}
}
