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 3×3 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.

这道题是之前那道 Unique Paths 不同的路径 的延伸，在路径中加了一些障碍物，还是用动态规划Dynamic Programming来解，不同的是当遇到为1的点，将该位置的dp数组中的值清零，其余和之前那道题并没有什么区别，代码如下：

解法一：

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

或者我们也可以使用一维dp数组来解，省一些空间，参见代码如下：

解法二：

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

类似题目：

Unique Paths