63. Unique Paths II(unsolved)

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]
]
The total number of unique paths is 2.

Note: m and n will be at most 100.

解答：这道题跟Unique Paths 解法是一样的，问题在于如何解决遇到障碍的问题。其实我们可以认为假如到了某个点（i，j），那么原来的解法是

dp[i][j]=dp[i-1][j]+dp[i][j-1]

而假如dp[i-1][j]这个位置是障碍，那么其实应认为它的方法是0，即
dp[i][j]无法从dp[i-1][j]过来。那么设定dp[i-1][j]=0；
即假如obstacleGrid[i][j]==1,那么设定dp[i][j]=0；

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