63 Unique Paths II (dp)

题目：

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 3   3 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.

同Unique Paths，加如了障碍。

方法一（dp记忆+深搜）：

代码：

public static int dp[][] = new int[100][100];
public int uniquePathsWithObstacles(int[][] obstacleGrid) {
int n= obstacleGrid.length;
int m=obstacleGrid[0].length;
if(obstacleGrid[0][0]==1 || obstacleGrid[n-1][m-1]==1)return 0;
dp[0][0]=1;
return dfs(n-1,m-1,obstacleGrid);
}

public int dfs(int x,int y,int[][] obstacleGrid)
{
if(x<0 || y<0)return 0;
if(obstacleGrid[x][y]==1)return 0;
if(x==0 && y==0)return dp[0][0];
if(dp[x][y]>0)return dp[x][y];
else
return dp[x][y]=dfs(x-1,y,obstacleGrid)+dfs(x,y-1,obstacleGrid);
}
方法二：
动态规划：

状态方程：f[i][j]=f[i-1][j]+f[i][j-1]

代码：

public int uniquePathsWithObstacles(int[][] obstacleGrid) {
int m=obstacleGrid.length;
int n=obstacleGrid[0].length;
int[] dp = new int
;

if(obstacleGrid[0][0] == 1 || obstacleGrid[m-1][n-1]==1)
{
return 0;
}
dp[0] = 1;
for(int i= 0;i<m;i++)
{
dp[0]=dp[0]==0?0:(obstacleGrid[i][0]==1?0:1);
for(int j=1;j<n;j++)
{
dp[j]=obstacleGrid[i][j]==1?0:(dp[j]+dp[j-1]);
}
}
return dp[n-1];
}