63. Unique Paths II(有障碍的路径 动态规划)
2024-08-29 12:43:18
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.
class Solution {
public int uniquePathsWithObstacles(int[][] obstacleGrid) {
int rows = obstacleGrid.length;
int cols = obstacleGrid[0].length;
int[][] dp = new int[rows][cols];
for(int i = 0; i < rows;i++){
for (int j = 0; j < cols ;j++){
if(obstacleGrid[i][j]==1)
dp[i][j] = 0;
else{
if (i==0&& j==0)
dp[i][j] = 1;
else if (i==0)
dp[i][j] = dp[i][j-1]; //边界
else if (j == 0)
dp[i][j] = dp[i-1][j];
else
dp[i][j] = dp[i-1][j] + dp[i][j-1];
}
}
}
return dp[rows-1][cols-1];
}
}
class Solution {
public int uniquePathsWithObstacles(int[][] obstacleGrid) {
int rows = obstacleGrid.length;
int cols = obstacleGrid[0].length;
for(int i = 0; i < rows;i++){
for (int j = 0; j < cols ;j++){
if(obstacleGrid[i][j]==1)
obstacleGrid[i][j] = 0;
else if (i==0&& j==0)
obstacleGrid[i][j] = 1;
else if (i==0)
obstacleGrid[i][j] = obstacleGrid[i][j-1]*1; //边界，没有路径了，要么是0，要么是1
else if (j == 0)
obstacleGrid[i][j] = obstacleGrid[i-1][j]*1;
else
obstacleGrid[i][j] = obstacleGrid[i-1][j] + obstacleGrid[i][j-1];
}
}
return obstacleGrid[rows-1][cols-1];
}
}
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