63.不同路径2

链接：

63.不同路径2

思路：

对于有障碍的节点map[i][j]=1，dp[i][j]=0。对于无障碍节点map[i][j]=0，若map[i-1][j]=0，则dp[i][j]+=dp[i-1][j]；若map[i][j-1]=0，则dp[i][j]+=dp[i][j-1]；

实现：

class Solution {
public:
    int uniquePathsWithObstacles(vector<vector<int>> &obstacleGrid) {
        int m = obstacleGrid.size();
        vector<int> row = obstacleGrid.at(0);
        int n = row.size();
        int path[110][110];
        memset(path, 0, sizeof(path));
        int cnt = 0;
        if (obstacleGrid[0][0] == 0) {
            path[0][0] = 1;
        }
        for (int i = 1; i < m; i++) {
            if (obstacleGrid[i][0] == 1) {
                path[i][0] = 0;
                continue;
            }
            if (obstacleGrid[i - 1][0] == 0) {
                path[i][0] += path[i - 1][0];
            }
        }
        for (int j = 1; j < n; j++) {
            if (obstacleGrid[0][j] == 1) {
                path[0][j] = 0;
                continue;
            }
            if (obstacleGrid[0][j - 1] == 0) {
                path[0][j] += path[0][j - 1];
            }
        }
        for (int i = 1; i < m; i++) {
            for (int j = 1; j < n; j++) {
                if (obstacleGrid[i][j] == 1) {
                    path[i][j] = 0;
                    continue;
                }
                if (obstacleGrid.at(i - 1).at(j) == 0) {
                    path[i][j] += path[i - 1][j];
                }
                if (obstacleGrid.at(i).at(j - 1) == 0) {
                    path[i][j] += path[i][j - 1];
                }
            }
        }
        return path[m - 1][n - 1];
    }
};