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
   Accepted
   266,240
   Submissions
   784,147

注意long long

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