【leetcode】63. Unique Paths II

题目：

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思路：
思路与62. Unique Paths一样，用动态规划，比较简单。唯一的区别是，这里把障碍所在位置的dp[i][j]设置成0即可。

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代码实现：
注意：dp数组之所以用long long的形式是因为在相加时会发生signed int溢出。

class Solution {
public:
    int uniquePathsWithObstacles(vector<vector<int>>& obstacleGrid) {
        long long dp[100][100];
        int rows = obstacleGrid.size();
        int cols = obstacleGrid[0].size();
        
        bool has_obs = false;
        for (int i = 0; i < rows; ++i){
            if (obstacleGrid[i][0] == 1){
                has_obs = true;
            }
            if (has_obs == true){
                dp[i][0] = 0;
            }else{
                dp[i][0] = 1;
            }
        }
        has_obs = false;
        for (int i = 0; i < cols; ++i){
            if (obstacleGrid[0][i] == 1){
                has_obs = true;
            }
            if (has_obs == true){
                dp[0][i] = 0;
            }else{
                dp[0][i] = 1;
            }
        }
        
        for (int i = 1; i < rows; ++i){
            for (int j = 1; j < cols; ++j){
                dp[i][j] = 0;
                if (obstacleGrid[i][j] == 0){
                    if (i-1 >= 0){
                        dp[i][j] += dp[i-1][j];
                    }
                    if (j-1 >= 0){
                        dp[i][j] += dp[i][j-1];
                    }
                }
            }
        }
        
        return dp[rows-1][cols-1];
    }
};