【LeetCode】Unique Paths II

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.

java code : 一维的动态规划，遇到障碍设为0即可。

public class Solution {
public int uniquePathsWithObstacles(int[][] obstacleGrid) {
// IMPORTANT: Please reset any member data you declared, as
// the same Solution instance will be reused for each test case.
int row = obstacleGrid.length;
if(row == 0)
return 0;
int col = obstacleGrid[0].length;
if(col == 0)
return 0;
if(obstacleGrid[0][0] == 1 || obstacleGrid[row-1][col-1] == 1)
return 0;
int[][] dp = new int[row][col];
dp[0][0] = 1;
for(int i = 1; i < col; i++)
{
if(obstacleGrid[0][i] == 1)
dp[0][i] = 0;
else dp[0][i] = dp[0][i-1];
}
for(int i = 1; i < row; i++)
{
if(obstacleGrid[i][0] == 1)
dp[i][0] = 0;
else dp[i][0] = dp[i-1][0];
}
for(int i = 1; i < row; i++)
{
for(int j = 1; j < col; j++)
{
if(obstacleGrid[i][j] == 1)
dp[i][j] = 0;
else
{
dp[i][j] = dp[i][j-1] + dp[i-1][j];
}
}
}
int res = dp[row-1][col-1];
for(int i = 0; i < row; i++)
dp[i] = null;
dp = null;
return res;
}
}