**一. 题目描述** 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. **二. 题目分析** 与上一题Unique Paths类似，但要特别注意第一列的障碍。在上一题中，第一列全部是1，但是在这一题中不同的是，第一列如果某一行有障碍物，那么后面的行应该为0。 **三. 示例代码** 使用动态规划： ~~~ #include <iostream> #include <vector> using namespace std; class Solution { public: int uniquePathsWithObstacles(vector<vector<int> > &obstacleGrid) { const size_t x = obstacleGrid.size(); // 行数 const size_t y = obstacleGrid[0].size(); // 列数 vector<vector<int> > k; for (int i = 0; i < x; ++i) k.push_back(vector<int>(y, 0)); for (int i = 0; i < x; ++i) { if (obstacleGrid[i][0] == 0) k[i][0] = 1; else { for (int p = i; p < x; ++p) k[p][0] = 0; break; } } for (int j = 0; j < y; ++j) { if (obstacleGrid[0][j] == 0) k[0][j] = 1; else { for (int q = j; q < y; ++q) k[0][q] = 0; break; } } for (int i = 1; i < x; ++i) { for (int j = 1; j < y; ++j) { if (obstacleGrid[i][j] != 0) k[i][j] = 0; else k[i][j] = k[i - 1][j] + k[i][j - 1]; } } return k[x - 1][y - 1]; } }; ~~~  **四. 小结** [](http://blog.csdn.net/liyuefeilong/article/details/49520819#)[](http://blog.csdn.net/liyuefeilong/article/details/49520819# "分享到QQ空间")[](http://blog.csdn.net/liyuefeilong/article/details/49520819# "分享到新浪微博")[](http://blog.csdn.net/liyuefeilong/article/details/49520819# "分享到腾讯微博")[](http://blog.csdn.net/liyuefeilong/article/details/49520819# "分享到人人网")[](http://blog.csdn.net/liyuefeilong/article/details/49520819# "分享到微信")