Unique Paths II

标签：算法

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];
    }
}

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];
    }
}

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);
    }
}

Unique Paths II,布布扣,bubuko.com

Unique Paths II

标签：算法

原文地址：http://blog.csdn.net/dutsoft/article/details/38638757