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.
题目标签：Array
题目给了我们一个matrix，里面1代表有障碍物，0代表空。这道题目和Unique Path 基本一样。之前我们是给start point 1的值，然后遍历matrix，拿left 和 top的值加起来。那么这题，1被用了，代表障碍物。那我们就用-1。首先来分析一下，如果起点或者终点都是1的话，那么就无路可走，直接return 0就可以。剩下的情况，对于每一个点，如果它是障碍物，就直接跳过；如果不是，那么拿left 和top 的加一起，那如果left 和 top 有障碍物的话，也跳过。最后把终点的值 * -1 就可以了。
当然，也可以另外新建一个matrix，过程是同样的方法，不同的是当这个点是1的话，就把这个点的值等于0。 这方法就需要多建一个matrix。
Java Solution:
Runtime beats 14.83%
完成日期：07/21/2017
关键词：Array
关键点：Dynamic Programming， 逆向思考
1 public class Solution
2 {
3 public int uniquePathsWithObstacles(int[][] obstacleGrid)
4 {
5 if(obstacleGrid[0][0] == 1 ||
6 obstacleGrid[obstacleGrid.length-1][obstacleGrid[0].length-1] == 1) // obstacle at start point or finish point
7 return 0;
8
9 obstacleGrid[0][0] = -1; // start point
10
11 for(int i=0; i<obstacleGrid.length; i++) // row
12 {
13 for(int j=0; j<obstacleGrid[0].length; j++) // column
14 {
15 // if this is not obstacle
16 if(obstacleGrid[i][j] !=1)
17 {
18 // get left: left is not obstacle
19 if(j-1 >=0 && obstacleGrid[i][j-1] !=1)
20 obstacleGrid[i][j] += obstacleGrid[i][j-1];
21 // get top: top is not obstacle
22 if(i-1 >=0 && obstacleGrid[i-1][j] !=1)
23 obstacleGrid[i][j] += obstacleGrid[i-1][j];
24 }
25
26 }
27 }
28
29 return obstacleGrid[obstacleGrid.length-1][obstacleGrid[0].length-1] * -1;
30 }
31 }
参考资料：N/A
LeetCode 算法题目列表 - LeetCode Algorithms Questions List