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.
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]; } };
本文转自博客园Grandyang的博客，原文链接：不同的路径之二[LeetCode] Unique Paths II ，如需转载请自行联系原博主。