63. Unique Paths II

题意：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.

思路：这题和之前的62. Unique Paths差不多，就多了个障碍点。具体思路还是一样，但是这里需要把障碍点的结果值置为0，所以需要做一些修改：

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

我的方法利用几个if判断来对第一行和第一列的结果进行一个选择，如果之前有障碍，则第一行/列之后的都为0，当然，后来看到更简单的方法，直接多开一行和一列的空间，全赋0，和62题的做法就基本一致了：

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]
;
}
};