LeetCode OJ算法题（六十二）：Unique Paths II

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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 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.

解法：

这里可以利用Unique Path的第二种想法，用一张（m+1）*（n+1）的表result存储到达某点时的路径数，仍然有result[i][j] = result[i-1][j] + result[i][j-1]。唯一需要注意的是，在原始矩阵中为1的位置所对应的result应该为0。且在边界处，只要出现了1，则后面（含该点）所有的result值都应该为0。

public class No62_UniquePathsII {public static void main(String[] args){System.out.println(uniquePathsWithObstacles(new int[][]{{1,0}}));}public static int uniquePathsWithObstacles(int[][] obstacleGrid) {int m = obstacleGrid.length;if(m == 0) return 0;int n = obstacleGrid[0].length;        int[][] result = new int[m][n];        int i=0,j=0;        for(;i<m;i++){        if(obstacleGrid[i][0]==1){        break;        }        else        result[i][0] = 1;        }        while(i<m) result[i++][0] = 0;        for(;j<n;j++){        if(obstacleGrid[0][j]==1) break;        else result[0][j] = 1;        }        while(j<n) result[0][j++] = 0;        for(i=1;i<m;i++){        for(j=1;j<n;j++){        if(obstacleGrid[i][j] != 1)        result[i][j] = result[i-1][j] + result[i][j-1];        else         result[i][j] = 0;        }        }        return result[m-1][n-1];    }}

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