Unique Paths II

思路和Unique paths 差不多，但是由于有了障碍物，所以，初始化dp矩阵的时候，初始化为0___________________________________________________


class Solution {
public:
    int uniquePathsWithObstacles(vector<vector<int>>& obstacleGrid) {
        if (obstacleGrid.empty() || obstacleGrid[0].empty() || obstacleGrid[0][0] == 1) return 0;
        int m=obstacleGrid[0].size(),n=obstacleGrid.size();
        vector<vector<int>> path(n,vector<int>(m,0));
        path[0][0]=1;
        for(int i=0;i<n;i++)
            for(int j=0;j<m;j++){
                 if(i==0&&j>0)path[i][j]=path[i][j-1];//第一次做的时候做错了，没有考虑到障碍物把路都封死的情况，写成了path[i][j]=1;
                if(j==0&&i>0)path[i][j]=path[i-1][j];
                if(i>0&&j>0)path[i][j]=path[i-1][j]+path[i][j-1];
                if(obstacleGrid[i][j]==1)path[i][j]=0;
            }
        return path[n-1][m-1];
    }
};

Runtime: 4 ms, faster than 52.68% of C++ online submissions for Unique Paths II.速度比较慢，看看大佬的写法

看讨论区说这个是0ms，然而我运行还是4ms，23333

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