LeetCode63. Unique Paths II（C++）

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

解题思路：用一个cnt数组记录从出发点到该点的路径数，如果该点有障碍，则该点的cnt为0，若该点无障碍，则该点的cnt为该点上面一点的cnt与左边一点的cnt的和，同时要考虑行列号为0的情况。

class Solution {
public:
    int uniquePathsWithObstacles(vector>& obstacleGrid) {
        int m=obstacleGrid.size(),n=obstacleGrid[0].size();
        if(m == 0||n == 0)
            return 0;
        int cnt[m][n]={0};
        for(int i = 0; i < m; ++i){
            for(int j = 0; j < n; ++j){
                if(obstacleGrid[i][j] == 1)
                    cnt[i][j] = 0;
                else if (i == 0&&j == 0)
                    cnt[i][j] = 1;
                else if(i == 0)
                    cnt[i][j] = cnt[i][j-1];
                else if(j == 0)
                    cnt[i][j] = cnt[i-1][j];
                else
                    cnt[i][j] = cnt[i][j-1] + cnt[i-1][j];
            }
                
        }
        return cnt[m-1][n-1];
    }
    
};