Unique Paths II

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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.

dp解法，注意第一行或第一列有一个为1，则该行或该列后到达路径全为0.

public class Solution {    public int uniquePathsWithObstacles(int[][] obstacleGrid) {        if(obstacleGrid==null || obstacleGrid[0][0]==1) return 0;        int m=obstacleGrid.length;        int n=obstacleGrid[0].length;        int [][]f=new int[m][n];        for(int i=0;i<n;i++){            if(obstacleGrid[0][i]==0) f[0][i]=1;            else{                for(int k=i;k<n;k++)                    f[0][k]=0;                break;            }         }         for(int j=0;j<m;j++){            if(obstacleGrid[j][0]==0) f[j][0]=1;            else{                for(int k=j;k<m;k++)                    f[k][0]=0;                break;            }         }         for(int i=1;i<m;i++){            for(int j=1;j<n;j++){                if(obstacleGrid[i][j]==1) f[i][j]=0;                else{                    f[i][j]=f[i-1][j]+f[i][j-1];                }            }        }        return f[m-1][n-1];    }}

滚动数组dp

public class Solution {    public int uniquePathsWithObstacles(int[][] obstacleGrid) {        if(obstacleGrid==null || obstacleGrid[0][0]==1) return 0;        int m=obstacleGrid.length;        int n=obstacleGrid[0].length;        int []f=new int[n];        for(int i=0;i<n;i++){            if(obstacleGrid[0][i]==0) f[i]=1;            else{                for(int k=i;k<n;k++)                    f[k]=0;                break;            }        }        for(int i=1;i<m;i++){            for(int j=0;j<n;j++){                if(j==0 ){                    if(obstacleGrid[i][j]==1){                        f[j]=0;                    }                     continue;                }                if(obstacleGrid[i][j]==1) f[j]=0;                else{                    f[j]=f[j]+f[j-1];                }            }        }        return f[n-1];    }}

dfs

public class Solution {    public int [][]f;    public int uniquePathsWithObstacles(int[][] obstacleGrid) {        if(obstacleGrid==null || obstacleGrid[0][0]==1) return 0;        int m=obstacleGrid.length;        int n=obstacleGrid[0].length;        f=new int[m+1][n+1];        return dfs(m,n,obstacleGrid);    }    public int dfs(int m,int n,int[][] obstacleGrid){        if(m<1 || n<1 || obstacleGrid[m-1][n-1]==1) return 0;//        if(m==1 && n==1) return 1;//reach start point        return record(m-1,n,obstacleGrid)+record(m,n-1,obstacleGrid);//    }    public int record(int x,int y,int[][] obstacleGrid){        if(f[x][y]>0) return f[x][y];        else return f[x][y]=dfs(x,y,obstacleGrid);    }}

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