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：
与62. Unique Paths有以下不同：
1. 当(i, j)有障碍时dp[i][j] = 0
2. dp[0][j]和dp[i][0]未必为1.
dp[0][j] = obstacleGrid[0][j] ? 0 : dp[0][j-1]
dp[i][0] = obstacleGrid[i][0] ? 0 : dp[i-1][0]
3. 当obstacleGrid [0][0] = 1时，return 0
Java:
class Solution {
public int uniquePathsWithObstacles(int[][] obstacleGrid) {
int m = obstacleGrid.length;
int n = obstacleGrid[0].length;
if (m == 0 || n == 0) {
return 0;
}
if (obstacleGrid[0][0] == 1 || obstacleGrid[m-1][n-1] == 1) {
return 0;
}
int[][] dp = new int[m][n];
dp[0][0] = 1;
for(int i = 1; i < n; i++){
if(obstacleGrid[0][i] == 1)
dp[0][i] = 0;
else
dp[0][i] = dp[0][i-1];
}
for(int i = 1; i < m; i++){
if(obstacleGrid[i][0] == 1)
dp[i][0] = 0;
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][j-1] + dp[i-1][j];
}
}
return dp[m-1][n-1];
}
}
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