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.
这道题和上面那道只是多了一个障碍物的选择问题
思路是一样的，递推关系式也是一样的paths[i][j] = paths[i - 1][j] + paths[i][j - 1]。注意控制边界条件，和有障碍物时候paths的计算。
还要注意paths[i][j]和obstacleGrid[][]下标的对应关系，搞清楚了对应关系做起来就事半功倍了
1 public class Solution { 2 public int uniquePathsWithObstacles(int[][] obstacleGrid) { 3 int paths[][] = new int[obstacleGrid.length + 1][obstacleGrid[0].length + 1]; 4 if(obstacleGrid[0][0] == 1) 5 paths[1][1] = 0; 6 else 7 paths[1][1] = 1; 8 9 for(int i = 1; i < paths.length; i++){ 10 for(int j = 1; j < paths[0].length; j++){ 11 if(i == 1 && j != 1){ 12 if(obstacleGrid[i - 1][j - 1] != 1) 13 paths[i][j] = paths[i][j - 1]; 14 else 15 paths[i][j] = 0; 16 }//if 17 if(j == 1 && i != 1){ 18 if(obstacleGrid[i - 1][j - 1] != 1) 19 paths[i][j] = paths[i - 1][j]; 20 else 21 paths[i][j] = 0; 22 }//if 23 if(i != 1 && j != 1){ 24 if(obstacleGrid[i - 1][j - 1] == 1) 25 paths[i][j] = 0; 26 else 27 paths[i][j] = paths[i - 1][j] + paths[i][j - 1]; 28 } 29 }//for 30 }//for 31 32 return paths[obstacleGrid.length][obstacleGrid[0].length]; 33 } 34 }
原文：http://www.cnblogs.com/luckygxf/p/4234841.html