63. Unique Paths II

做题历程：

1. 应该不止2次做这道题了，大概耗时17分钟，独立解出

本题的难度其实算是比较低的，跟62. Unique Paths基本是异曲同工，解法也是参照那道题，使用dynamic programming，唯一注意的是遇到obstacle的处理，需要置零，因为该点永远不可能到达。 代码如下：

class Solution {
public:
    int uniquePathsWithObstacles(vector<vector<int>>& obstacleGrid) {
        if (obstacleGrid.empty() || obstacleGrid[0].empty()) {
            return 0;
        }
        int m = obstacleGrid.size();
        int n = obstacleGrid[0].size();
        vector<vector<int>> dp(m, vector<int>(n, 0));
        for (int i = 0; i < n; ++i) {
            if (obstacleGrid[0][i] == 1) {
                break;
            }
            dp[0][i] = 1;
        }

        for (int i = 0; i < m; ++i) {
            if (obstacleGrid[i][0] == 1) {
                break;
            }
            dp[i][0] = 1;
        }

        for (int i = 1; i < m; ++i) {
            for (int j = 1; j < n; ++j) {
                if (obstacleGrid[i][j] == 1) {
                    continue;
                }
                dp[i][j] = dp[i-1][j] + dp[i][j-1];
            }
        }
        return dp[m-1][n-1];
    }
};