leetcode 63. Unique Paths II

/*leetcode 63. Unique Paths IIFollow 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.解题思路：动态规划dp[j] = dp[j] + dp[j-1]*/#include <iostream>#include <vector>#include <string>using namespace std;class Solution {public:    int uniquePathsWithObstacles(vector<vector<int>>& obstacleGrid)     {        int m = obstacleGrid.size();        int n = obstacleGrid[0].size();        vector<int> dp(n ,0);   //初始化都为0        dp[0] = 1;      //第0行0列的步数        for (int i = 0; i < m; ++i)        {            if (obstacleGrid[i][0] == 1)        //边界处理：有一个为1，说明此路不通                dp[0] = 0;            for (int j = 1; j < n; ++j)            {                //dp[j]为第i行，第j列的步数                if (obstacleGrid[i][j] == 1)                    dp[j] = 0;                else                    dp[j] = dp[j] + dp[j - 1];                cout << i << "," << j << ":" << dp[j] << endl;            }        }        return dp[n - 1];    }};void test(){    vector<vector<int>> ob{        {0,0,0},        {0,1,0},        {0,0,0}    };    Solution sol;    cout << sol.uniquePathsWithObstacles(ob) << endl;}int main(){    test();    return 0;}