【动态规划】63. 不同路径 II

链接：63. 不同路径 II - 力扣（LeetCode）

问题描述：

解释：障碍物的位置对路径总数的贡献为0

    int uniquePathsWithObstacles(vector>& obstacleGrid) {
        int m = obstacleGrid.size();
        int n = obstacleGrid[0].size();
        vector> dp(m, vector(n, 0));
        for (int i = 0; i < m; i++) {
            if (obstacleGrid[i][0] == 1) break;
            dp[i][0] = 1;
        }
        for (int j = 0; j < n; j++) {
            if (obstacleGrid[0][j] == 1) break;
            dp[0][j] = 1;
        }
        for (int i = 1; i < m; i++) {
            for (int j = 1; j < n; j++) {
                if (obstacleGrid[i][j] == 1) {
                    dp[i][j] = 0;
                    continue;
                }
                dp[i][j] = dp[i-1][j] + dp[i][j-1];
            }
        }
        return dp[m-1][n-1];
    }