LeetCode OJ：Unique Paths II（唯一路径II）

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]
]

这个与前面那个题目基本上差不多，具体就是多了障碍，实际上遇到障碍的话把到该点的路径的数目置为0就可以了，其他的基本上与前面相同：

 class Solution {
 public:
     int uniquePathsWithObstacles(vector<vector<int>>& obstacleGrid) {
         if(obstacleGrid.size() ==  || obstacleGrid[].size() == ) return ;
         vector<vector<int>> res(obstacleGrid.size(), vector<int>(obstacleGrid[].size(), ));
         int maxHor = obstacleGrid.size();
         int maxVer = obstacleGrid[].size();
         res[][] = obstacleGrid[][] ==  ?  : ;
         for(int i = ; i < maxHor; ++i){
             res[i][] = obstacleGrid[i][] ==  ?  : res[i - ][];
         }
         for(int j = ; j < maxVer; ++j){
             res[][j] = obstacleGrid[][j] ==  ?  : res[][j - ];
         }
         for(int i = ; i < maxHor; ++i){
             for(int j = ; j < maxVer; ++j){
                 res[i][j] = obstacleGrid[i][j] ==  ?  : res[i - ][j] + res[i][j - ];
             }
         }
         return res[maxHor - ][maxVer - ];
     }
 };

java版本的代码如下所示：

 public class Solution {
     public int uniquePathsWithObstacles(int[][] obstacleGrid) {
         if(obstacleGrid.length == 0 || obstacleGrid[0].length == 0)
             return 0;
         int [][] grid = new int [obstacleGrid.length][obstacleGrid[0].length];
         grid[0][0] = obstacleGrid[0][0]==1 ? 0 : 1;
         for(int i = 1; i < obstacleGrid.length; ++i){
             grid[i][0] = obstacleGrid[i][0]==1? 0 : grid[i-1][0];
         }
         for(int i = 1; i < obstacleGrid[0].length; ++i){
             grid[0][i] = obstacleGrid[0][i]==1? 0 : grid[0][i-1];
         }
         for(int i = 1; i < obstacleGrid.length; ++i){
             for(int j = 1; j < obstacleGrid[0].length; ++j){
                 grid[i][j] = obstacleGrid[i][j]==1? 0 : (grid[i-1][j] + grid[i][j-1]);
             }
         }
         return grid[obstacleGrid.length-1][obstacleGrid[0].length-1];
     }
 }