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.
这道题就是上一道题的延伸,相当于上一道题就是给定了一个m*n的矩阵,而且全是0,这里位置1表示的不能走的地方。所以在上一个版本上修正。
区别就是在于如果遇到1的时候的处理方式不同。
public class Solution { public int uniquePathsWithObstacles(int[][] obstacleGrid) { int m = obstacleGrid.length; int n = obstacleGrid[0].length; if( m == 0 || n == 0) return 0; if( m == 1){ for( int i = 0;i<n;i++){ if( obstacleGrid[0][i] == 1 ) return 0; } return 1; } if( n == 1){ for( int i = 0;i<m;i++){ if( obstacleGrid[i][0] == 1) return 0; } return 1; } int[] dp = new int[n]; for( int i = 0;i<n;i++){ if( obstacleGrid[0][i] == 1 ) while(i<n){ dp[i] = 0; i++; } else dp[i] = 1; } for( int i = 1;i < m;i++){ if( obstacleGrid[i][0] == 1){ int j = 0; while(j<n){ if( obstacleGrid[i][j] == 1) dp[j] = 0; else dp[j] = dp[j]; j++; } } for( int j = 1;j<n;j++){ if( obstacleGrid[i][j] == 1){ dp[j] = 0; }else dp[j] += dp[j-1]; } } return dp[n-1]; } }