leetcode Unique Paths II

题目链接

这里有个效率很低的算法。深度优先算法

public class Solution {

    public static void main(String args[])
    {
        Solution sol=new Solution();
        System.out.println(sol.uniquePathsWithObstacles(new int[][]{{0,0,0,0,0,0,0,0,1,0,1,0,0,0,0,0,0,0,0,1,0,0,0,0,0,1,0,0,0,0,0,0,0},{0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1},{0,0,0,0,0,1,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,1,0,0,0,0,0,0},{1,1,1,0,0,0,0,1,0,0,0,0,0,0,0,0,1,0,0,1,1,0,0,0,0,0,0,0,0,1,0,0,1},{0,0,1,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,1,0,0,0,0,0},{0,0,0,1,0,1,0,0,0,0,1,1,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,1,1,0},{1,0,1,1,1,0,0,0,0,1,0,0,0,0,0,0,0,0,1,0,0,0,0,0,1,0,0,0,0,0,0,0,0},{0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,1,0,1,0,0,0,1,0,1,0,0,0,0,0,0},{0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,1,0,0,1,0},{0,0,0,0,0,0,0,0,0,1,1,0,0,0,0,0,0,0,0,0,0,0,1,1,0,0,0,1,0,0,0,0,0},{0,1,1,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0},{1,0,1,0,0,0,0,0,0,0,0,1,0,0,1,0,0,0,0,1,0,1,0,0,0,1,0,1,0,0,0,0,1},{0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,1,0,1,0,0,0,0,0,0,1,1,0,0,0,0,0},{0,1,0,1,0,0,0,0,1,0,0,1,0,0,0,0,0,0,0,1,1,0,0,0,0,0,0,1,0,0,0,0,0},{0,1,0,0,0,0,0,0,1,0,0,1,1,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,1,1,0,1},{1,0,0,0,0,1,0,0,1,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0},{0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,1,0,0,0,0,1,1,0,0,1,0,0,0,0,0,0},{0,0,1,0,0,0,0,0,0,0,1,0,0,1,0,0,1,0,0,0,0,0,0,1,1,0,1,0,0,0,0,1,1},{0,1,0,0,1,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,1,0,1,1,0,1,0,1},{1,1,1,0,1,0,0,0,0,1,0,0,0,0,0,0,1,0,1,0,1,1,0,0,0,0,0,0,0,0,0,0,0},{0,0,0,0,1,1,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,1,0,0,0,0,0,0,0,0,0,1,1},{0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,1,0,1,0,0,0,1,0,0,0}}));

    }


    int count;
    int [][] obstacleGrid;
    int m;
    int n;
    public int uniquePathsWithObstacles(int[][] obstacleGrid) {
        this.obstacleGrid=obstacleGrid;
        m=obstacleGrid.length;
        n=obstacleGrid[0].length;
        dfs(0,0);
        return count;
    }

    public void dfs(int i,int j)
    {
        if(i>=m||j>=n) //超界
        {
            return ;
        }

        if(i==m-1&&j==n-1)//终点
        {
            count++;
            return;
        }
        if(obstacleGrid[i][j]==1)//撞墙
        {
            return;
        }   
        dfs(i+1,j);
        dfs(i,j+1);
    }


}

当然还有更好的基于dp的算法。

record[i][j]记录的是从起点到达ij的所有可能的步数。如果ij是不能到达的记为0.

int count;
    int [][] obstacleGrid;
    int [][] record;
    int m;
    int n;
    public int uniquePathsWithObstacles(int[][] obstacleGrid) {
        this.obstacleGrid=obstacleGrid;
        m=obstacleGrid.length;
        n=obstacleGrid[0].length;
        record=new int[m][n];
        record[0][0]=1;
        for(int i=0;i<m;i++)
        {
            for(int j=0;j<n;j++)
            {
                dp(i,j);
            }
        }
        return record[m-1][n-1];
    }

    public void dp(int i,int j)
    {
        if(obstacleGrid[i][j]!=1)
        {
            if(i-1>=0)
            {
                record[i][j]+=record[i-1][j];
            }
            if(j-1>=0)
            {
                record[i][j]+=record[i][j-1];
            }
        }
        else
        {
            record[i][j]=0;
        }
    }