Unique Path II

题目：

Follow 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.

分析：

题目本身并不难，会做Unique Path就一定能会做Unique Path II.

解法：

class Solution {
public:
    /**
     * @param obstacleGrid: A list of lists of integers
     * @return: An integer
     */ 
    int uniquePathsWithObstacles(vector<vector<int> > &obstacleGrid) {
        // write your code here
        if (obstacleGrid.size() == 0) {
            return 0;
        }

        if (obstacleGrid[0].size() == 0) {
            return 0;
        }

        if (obstacleGrid[0][0] == 1) {
            return 0;
        }

        int m = obstacleGrid.size();
        int n = obstacleGrid[0].size();

        vector<vector<int> > res(m, vector<int>(n, 0) );

        res[0][0] = 1;
        for (int i = 1; i < m; i++) {
            if (obstacleGrid[i][0] == 1) {
                res[i][0] = 0;
            } else {
                res[i][0] = res[i - 1][0];
            }
        }

        for (int j = 1; j < n; j++) {
            if (obstacleGrid[0][j] == 1) {
                res[0][j] = 0;
            } else {
                res[0][j] = res[0][j - 1];
            }
        }

        for (int i = 1; i < m; i++) {
            for (int j = 1; j < n; j++) {
                if(obstacleGrid[i][j] == 1) {
                    res[i][j] = 0;
                } else {
                    res[i][j] = res[i - 1][j] + res[i][j - 1];
                }
            }
        }

        return res[m - 1][n - 1];
    }
};