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A robot is located at the top-left corner of a m x n grid (marked 'Start' in the diagram below).
The robot can only move either down or right at any point in time. The robot is trying to reach the bottom-right corner of the grid (marked 'Finish' in the diagram below).
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.
Note: m and n will be at most 100.
Example 1:
Input: [ [0,0,0], [0,1,0], [0,0,0] ] Output: 2 Explanation: There is one obstacle in the middle of the 3x3 grid above. There are two ways to reach the bottom-right corner: 1. Right -> Right -> Down -> Down 2. Down -> Down -> Right -> Right
使用动态规划的方式，二维数组表示到当前位置可以有的路径数量，其确定需要根据当前位置的左边格子和上边格子的值，如果当前位置被挡住，即obstacleGrid[i][j]的当前元素值为1，则dp[i][j]=0，否则dp[i][j] = dp[i-1][j] + dp[i][j-1]。同时关键的是判断边界条件，第一行和第一列只看上边或者左边的值就可以确定当前值。
class Solution {
public int uniquePathsWithObstacles(int[][] obstacleGrid) {
int m = obstacleGrid.length;
if(m==0) return 0;
int n = obstacleGrid[0].length;
int[][] dp = new int[m][n];
dp[0][0] = obstacleGrid[0][0] == 1? 0:1;
for(int i=0; i<m; i++){
for(int j=0; j<n; j++){
if(i+j == 0) continue;
if(i == 0){
dp[i][j] = obstacleGrid[i][j] == 1? 0:dp[i][j-1];
}else if(j == 0){
dp[i][j] = obstacleGrid[i][j] == 1? 0:dp[i-1][j];
}else{
dp[i][j] = obstacleGrid[i][j] == 1? 0:dp[i-1][j] + dp[i][j-1];
}
}
}
return dp[m-1][n-1];
}
}