#115 Unique Paths II

题目描述：

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 0respectively in the grid.

 Notice

m and n will be at most 100.

Have you met this question in a real interview? 

Yes

Example

For example,

There is one obstacle in the middle of a 3x3 grid as illustrated below.

[  [0,0,0],  [0,1,0],  [0,0,0]]

The total number of unique paths is 2.

题目思路：

这题和Unique Paths类似，不同点在于多加了障碍的情况。dp方程还是一样的，只是多了一个条件：如果当前点有障碍，那么dp[i][j] ＝ 0。

Mycode （AC ＝ 59ms）：

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;        vector<vector<int>> dp(obstacleGrid);                // intialize the first row and first column        dp[0][0] = obstacleGrid[0][0] == 1? 0 : 1;        for (int i = 1; i < obstacleGrid.size(); i++) {            if (obstacleGrid[i][0] == 0 && dp[i - 1][0] == 1) {                dp[i][0] = 1;            }            else {                dp[i][0] = 0;            }        }                for (int j = 1; j < obstacleGrid[0].size(); j++) {            if (obstacleGrid[0][j] == 0 && dp[0][j - 1] == 1) {                dp[0][j] = 1;            }            else {                dp[0][j] = 0;            }        }                // do dp: if no obstacle,         // then dp[i][j] = dp[i - 1][j] + dp[i][j - 1]        for (int i = 1; i < obstacleGrid.size(); i++) {            for (int j = 1; j < obstacleGrid[0].size(); j++) {                if (obstacleGrid[i][j] == 0) {                    dp[i][j] = dp[i - 1][j] + dp[i][j - 1];                }                else {                    dp[i][j] = 0;                }            }        }                return dp[obstacleGrid.size() - 1][obstacleGrid[0].size() - 1];    }};