题目描述：一个机器人位于一个 m x n 网格的左上角 （起始点在下图中标记为“Start” ）。
机器人每次只能向下或者向右移动一步。机器人试图达到网格的右下角（在下图中标记为“Finish”）。
现在考虑网格中有障碍物。那么从左上角到右下角将会有多少条不同的路径？
示例：输入:
[
[0,0,0],
[0,1,0],
[0,0,0]
]
输出: 2
解释:
3x3 网格的正中间有一个障碍物。
从左上角到右下角一共有 2 条不同的路径：
1. 向右 -> 向右 -> 向下 -> 向下
2. 向下 -> 向下 -> 向右 -> 向右
代码：
java：
class Solution {
public int uniquePathsWithObstacles(int[][] obstacleGrid) {
int R = obstacleGrid.length;
int C = obstacleGrid[0].length;
if(obstacleGrid[0][0] == 1) return 0;
obstacleGrid[0][0] = 1;
for(int i=1;i<R;i++) {
obstacleGrid[i][0] = (obstacleGrid[i][0] == 0 && obstacleGrid[i-1][0] == 1)? 1 : 0;
}
for(int i=1;i<C;i++) {
obstacleGrid[0][i] = (obstacleGrid[0][i] == 0 && obstacleGrid[0][i-1] == 1)? 1 : 0;
}
for(int i=1;i<R;i++) {
for(int j=1;j<C;j++) {
if(obstacleGrid[i][j] == 0) {
obstacleGrid[i][j] = obstacleGrid[i-1][j] + obstacleGrid[i][j-1];
}else {
obstacleGrid[i][j] = 0;
}
}
}
return obstacleGrid[R-1][C-1];
}
}
python;
m = len(obstacleGrid)
n = len(obstacleGrid[0])
# If the starting cell has an obstacle, then simply return as there would be
# no paths to the destination.
if obstacleGrid[0][0] == 1:
return 0
# Number of ways of reaching the starting cell = 1.
obstacleGrid[0][0] = 1
# Filling the values for the first column
for i in range(1,m):
obstacleGrid[i][0] = int(obstacleGrid[i][0] == 0 and obstacleGrid[i-1][0] == 1)
# Filling the values for the first row
for j in range(1, n):
obstacleGrid[0][j] = int(obstacleGrid[0][j] == 0 and obstacleGrid[0][j-1] == 1)
# Starting from cell(1,1) fill up the values
# No. of ways of reaching cell[i][j] = cell[i - 1][j] + cell[i][j - 1]
# i.e. From above and left.
for i in range(1,m):
for j in range(1,n):
if obstacleGrid[i][j] == 0:
obstacleGrid[i][j] = obstacleGrid[i-1][j] + obstacleGrid[i][j-1]
else:
obstacleGrid[i][j] = 0
# Return value stored in rightmost bottommost cell. That is the destination.
return obstacleGrid[m-1][n-1]
作者：LeetCode
链接：https://leetcode-cn.com/problems/unique-paths-ii/solution/bu-tong-lu-jing-ii-by-leetcode/
来源：力扣（LeetCode）
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