Description：
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.
Code:
1 int uniquePathsWithObstacles(vector<vector<int>>& obstacleGrid) { 2 int m = obstacleGrid.size(); 3 int n = obstacleGrid[0].size(); 4 5 //path[i][j]表示(0,0)号元素到(i,j)号元素的路径数，所以最终结果为path[m-1][j-1] 6 7 int path[MAX][MAX] = {0}; 8 path[0][0] = (obstacleGrid[0][0] == 1)?0:1; 9 10 for (int i = 1; i < m; ++i) 11 { 12 path[i][0] = (obstacleGrid[i][0]==1)?0:path[i-1][0]; 13 } 14 15 for (int i = 1; i < n; ++i) 16 { 17 path[0][i] = (obstacleGrid[0][i]==1)?0:path[0][i-1]; 18 } 19 20 for (int i = 1; i < m; ++i) 21 { 22 for (int j = 1; j < n; ++j) 23 { 24 path[i][j] = (obstacleGrid[i][j] == 1)?0:(path[i-1][j] + path[i][j-1]); 25 } 26 } 27 return path[m-1][n-1]; 28 }
原文：http://www.cnblogs.com/happygirl-zjj/p/4582137.html