UniquePaths，UniquePaths2，路径问题。动态规划。

public class UniquePaths 
{
	//动态规划，非递归
	public int uniquePaths(int m, int n) 
	{
        int[][] a = new int[m][n];
        for(int i = 0; i < m; i ++)//初始条件，第一行第一列
        {
        	a[i][0] = 1;
        }
        for(int i = 0; i < n; i ++)
        {
        	a[0][i] = 1;
        }
        for(int i = 1; i < m; i ++)
        {
        	for(int j = 1; j < n; j ++)
        	{
        		a[i][j] = a[i-1][j]+a[i][j-1];//递推公式
        	}
        }
        return a[m-1][n-1];
    }
	
	//动态规划递归
	public int uniquePaths2(int m, int n)
	{
		if(m == 1 || n == 1) return 1;
		else
		{
			return uniquePaths2(m-1, n)+uniquePaths2(m, n-1);
		}
	}
}

public class UniquePaths2 
{
	public int uniquePathsWithObstacles(int[][] obstacleGrid) 
	{
        int m = obstacleGrid.length;
        int n = obstacleGrid[0].length;
        if(obstacleGrid[0][0] == 1 || obstacleGrid[m-1][n-1] == 1)//特例
        {
        	return 0;
        }
        for(int i = 0; i < m; i ++)//边界条件，第一行第一列，如果碰到1，则后面所有都为0
        {
        	if(obstacleGrid[i][0] == 1)
        	{
        		for(int j = i; j < m; j ++)
        		{
        			obstacleGrid[j][0] = 0;
        		}
        		break;
        	}
        	else
        	{
        		obstacleGrid[i][0] = 1;
        	}
        }
        for(int i = 1; i < n; i ++)
        {
        	if(obstacleGrid[0][i] == 1)
        	{
        		for(int j = i; j < n; j ++)
        		{
        			obstacleGrid[0][j] = 0;
        		}
        		break;
        	}
        	else
        	{
        		obstacleGrid[0][i] = 1;
        	}
        }
        
        for(int i = 1; i < m; i ++)
        {
        	for(int j = 1; j < n; j ++)
        	{
        		if(obstacleGrid[i][j] == 1)
        		{
        			obstacleGrid[i][j] = 0;
        		}
        		else
        		{
        			obstacleGrid[i][j] = obstacleGrid[i-1][j] + obstacleGrid[i][j-1];
        		}
        	}
        }
        return obstacleGrid[m-1][n-1];
    }
}