LeetCode-Unique Paths II

Unique Paths II

##题目

####Unique Paths II

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 3x3 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.

##解题思路
该题是Unique Paths的扩展，思路类似，只是这边要处理障碍。动态规划的思路可以见那题。

##算法代码
代码采用JAVA实现：

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public class Solution {
    public int uniquePathsWithObstacles(int[][] obstacleGrid) {
    	if(obstacleGrid==null || obstacleGrid.length==0)
    		return 0;
    	int m=obstacleGrid.length;
    	int n=obstacleGrid[0].length;
        int[][] dp=new int[m][n]; //dp[i][j]表示从start到[i,j]位置不同路径条数
        for(int i=0;i<m;i++)
        	for(int j=0;j<n;j++)
        		dp[i][j]=0;
        for(int i=0;i<n;i++)   //第一行障碍处理
        {
        	if(obstacleGrid[0][i]!=1)
        		dp[0][i]=1;
        	else
        		break;
        }
        	
        for(int j=0;j<m;j++)   //第一列障碍处理
        {
        	if(obstacleGrid[j][0]!=1)
        		dp[j][0]=1;
        	else
        		break;
        }
        for(int i=1;i<m;i++)
        	for(int j=1;j<n;j++)
        	{
        		if(obstacleGrid[i][j]==1)   //如果该位置是障碍，则到达该点的路径条数为0
        			dp[i][j]=0;
        		else
        			dp[i][j]=dp[i-1][j]+dp[i][j-1];
        	}		
        return dp[m-1][n-1];
    }
}

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