Leetcode 63 Unique Paths II

A robot is located at the top-left corner of a m x n grid (marked ‘Start’ in the diagram below).

The robot can only move either down or right at any point in time. The robot is trying to reach the bottom-right corner of the grid (marked ‘Finish’ in the diagram below).

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.

Note: m and n will be at most 100.

Example 1:

Input:
[
  [0,0,0],
  [0,1,0],
  [0,0,0]
]
Output: 2
Explanation:
There is one obstacle in the middle of the 3x3 grid above.
There are two ways to reach the bottom-right corner:
1. Right -> Right -> Down -> Down
2. Down -> Down -> Right -> Right

动态规划

和leetcode 62相同的方法. 只不过要加一个判断是不是obstacle的条件. 若(i,j)是obstacle, 那么(i,j)到终点的可能的路径数为0.

class Solution {
public:
    int uniquePathsWithObstacles(vector<vector<int>>& obstacleGrid) {
        vector<vector<int>> paths(obstacleGrid.size(), vector<int>(obstacleGrid.front().size(), -1));
        return uniquePathsWithObstacles(paths, obstacleGrid, 0, 0);
    }
private:
    int uniquePathsWithObstacles(vector<vector<int>>& paths, const vector<vector<int>>& obstacleGrid, int i, int j, int delimiterMark = -1, int obstacleMark = 1)
    {
        if(i == obstacleGrid.size() - 1 && j == obstacleGrid.front().size() - 1)
            return obstacleGrid[i][j] == obstacleMark ? 0 : 1;
        if(!isLocationValid(obstacleGrid, i, j))
            return 0;
        if(paths[i][j] != delimiterMark)
            return paths[i][j];
        int right = uniquePathsWithObstacles(paths, obstacleGrid, i, j+1);
        int down = uniquePathsWithObstacles(paths, obstacleGrid, i+1, j);
        paths[i][j] = right + down;
        return paths[i][j];
    }
    bool isLocationValid(const vector<vector<int>>& obstacleGrid, int i, int j, int delimiterMark = -1, int obstacleMark = 1)
    {
        return i < obstacleGrid.size() && j < obstacleGrid.front().size() && obstacleGrid[i][j] != obstacleMark;
    }
};