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() == 0 || obstacleGrid[0].size() == 0) return 0;
        vector<vector<int>> res(obstacleGrid.size(), vector<int>(obstacleGrid[0].size(), 0));
        int maxHor = obstacleGrid.size();
        int maxVer = obstacleGrid[0].size();
        res[0][0] = obstacleGrid[0][0] == 1 ? 0 : 1;
        for(int i = 1; i < maxHor; ++i){
            res[i][0] = obstacleGrid[i][0] == 1 ? 0 : res[i - 1][0];
        }
        for(int j = 1; j < maxVer; ++j){
            res[0][j] = obstacleGrid[0][j] == 1 ? 0 : res[0][j - 1];
        }
        for(int i = 1; i < maxHor; ++i){
            for(int j = 1; j < maxVer; ++j){
                res[i][j] = obstacleGrid[i][j] == 1 ? 0 : res[i - 1][j] + res[i][j - 1];
            }
        }
        return res[maxHor - 1][maxVer - 1];
    }
};

 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];
    }
}