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.

For example,

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

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

The total number of unique paths is 2.

Note: m and n will be at most 100.

思路：

只需要在
http://blog.csdn.net/lanxu_yy/article/details/17509431的基础上增加一个判断条件。

代码：

方法1：

class Solution {
public:
    int uniquePathsWithObstacles(vector<vector<int> > &obstacleGrid) {
        int m=obstacleGrid.size();
        int n = obstacleGrid[0].size();
        int ** dp=new int*[m];
        for(int i=0;i<m;i++)
        {
            dp[i] = new int[n];
        }
        
        int step = m+n-1;
        
        int cur=1;
        dp[0][0]=(obstacleGrid[0][0]==1?0:1);
        while(cur<step)
        {
            for(int i=0;i<=cur;i++)
            {
                if(i>=m)
                {
                    continue;
                }
                int j=cur-i;
                if(j>=n)
                {
                    continue;
                }
                if(obstacleGrid[i][j]==1)
                {
                    dp[i][j]=0;
                }
                else
                {
                    dp[i][j]=0;
                    if(i>0)
                    {
                        dp[i][j]+=dp[i-1][j];
                    }
                    if(j>0)
                    {
                        dp[i][j]+=dp[i][j-1];
                    }
                }
            }
            cur++;
        };
        return dp[m-1][n-1];
    }
};

方法2：

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