Leetcode 之 Unique Paths II

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.

 

和路径问题相似，采用动态规划的方法，从(0,0)到(m,n)仅允许向下和向右走，碰到1视为障碍。

C++代码如下：

class Solution {
public:
    int uniquePathsWithObstacles(vector<vector<int>>& obstacleGrid) {
        int m = obstacleGrid.size();
        int n = obstacleGrid[0].size();
        vector<vector<int> > a(m, vector<int>(n)); 

        int path1,path2,path=0;
        int j;
        for (int i = 0; i < m; i++) {  
            if(obstacleGrid[i][0] == 1)
                {
                    for(j = i; j< m ;j++)
                        a[j][0]=0;
                        break;
                }
            else
                        a[i][0]=1;
                
        }  
        for (int i = 0; i < n; i++) {  
            if(obstacleGrid[0][i] == 1)
                {
                    for(j = i; j< n ;j++)
                        a[0][j]=0;
                        break;
                }
            else
                        a[0][i]=1;
            
        }
        printf("%d",a[0][1]);        
        for (int i = 1; i < m; i++) {  
            for (int j = 1; j < n; j++) {  
                    path1=a[i-1][j]; 
                    path2=a[i][j-1]; 
                if(obstacleGrid[i][j])
                a[i][j]=0;
                else
                a[i][j]=path1+path2;
            }  
        }  
        return a[m-1][n-1]; 
    }  
        
};

 粘贴一个来源于别人http://blog.csdn.net/linhuanmars/article/details/22135231

更加简单的代码，以列或者行为扫描，每行扫描后更新新的一行所有点的路径数，再从左向右扫描，

public int uniquePathsWithObstacles(int[][] obstacleGrid) {  
    if(obstacleGrid == null || obstacleGrid.length==0 || obstacleGrid[0].length==0)  
        return 0;  
    int[] res = new int[obstacleGrid[0].length];  
    res[0] = 1;  
    for(int i=0;i<obstacleGrid.length;i++)  
    {  
        for(int j=0;j<obstacleGrid[0].length;j++)  
        {  
            if(obstacleGrid[i][j]==1)  
            {  
                res[j]=0;  
            }  
            else  
            {  
                if(j>0)  
                    res[j] += res[j-1];  
            }  
        }  
    }  
    return res[obstacleGrid[0].length-1];  
}  

好的代码是简洁的