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Unique Paths II

时间:2015-12-02 06:40:32      阅读:109      评论:0      收藏:0      [点我收藏+]

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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.

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.

Note

m and n will be at most 100.

 

 

public class Solution {
    /**
     * @param obstacleGrid: A list of lists of integers
     * @return: An integer
     */
    public int uniquePathsWithObstacles(int[][] obstacleGrid) {
        // write your code here
        
        if(obstacleGrid.length==0 || obstacleGrid[0].length==0)  return 0;
        int m=obstacleGrid.length;
        int n=obstacleGrid[0].length;
        
        
        int[][] sum=new int[m][n];
        int prev=0;
        
        for(int i=0;i<m;i++)
        {
            for(int j=0;j<n;j++)
            {
                
                if(obstacleGrid[i][j]==1)
                {
                    sum[i][j]=0;
                }
                else if(i==0 && j==0)
                {
                    sum[0][0]=1;
                }
                else if(i==0)
                {
                    sum[0][j]=sum[0][j-1];
                }
                else if(j==0)
                {
                    sum[i][0]=sum[i-1][0];
                }
                else
                {
                    sum[i][j]=sum[i-1][j]+sum[i][j-1];
                }
            }
        }
        
        return sum[m-1][n-1];
        
    }
}

 

Unique Paths II

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原文地址:http://www.cnblogs.com/kittyamin/p/5011865.html

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