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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.
Analyse: Construct another 2*2 array to store path information. First initialize the first column and first row, then compute remaining parts.
Runtime: 4ms.
1 class Solution { 2 public: 3 int uniquePathsWithObstacles(vector<vector<int>>& obstacleGrid) { 4 if(obstacleGrid[0][0] == 1) return 0; 5 6 int m = obstacleGrid.size(); 7 int n = obstacleGrid[0].size(); 8 9 vector<vector<int> > path(m, vector<int>(n, 0)); 10 path[0][0] = 1; 11 for(int i = 1; i < m; i++){ //indicate whether elements in the first column are reachable 12 if(obstacleGrid[i][0] == 0 && path[i - 1][0] == 1) path[i][0] = 1; 13 else path[i][0] = 0; 14 } 15 for(int j = 1; j < n; j++){ //indicate whether the elements in the first row are reachable 16 if(obstacleGrid[0][j] == 0 && path[0][j - 1] == 1) path[0][j] = 1; 17 else path[0][j] = 0; 18 } 19 for(int i = 1; i < m; i++){ 20 for(int j = 1; j < n; j++){ 21 if(obstacleGrid[i][j] == 1) path[i][j] = 0; 22 else path[i][j] = path[i - 1][j] + path[i][j - 1]; 23 } 24 } 25 return path[m - 1][n - 1]; 26 } 27 };
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原文地址：http://www.cnblogs.com/amazingzoe/p/4677438.html
