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
.
题目:这道题是上一题的升级版,中间加了一个障碍,其实只要把障碍地方的解变为 0 就好。
public class Solution {
public int uniquePathsWithObstacles(int[][] obstacleGrid) {
int row = obstacleGrid.length;
if(row < 1 ) return 0;
int col = obstacleGrid[0].length;
if(col < 1) return 0;
if(obstacleGrid[0][0] == 1|| obstacleGrid[row-1][col-1]==1) return 0;
for(int i = 0;i < row;i++) {
for(int j = 0;j < col;j++) {
if(obstacleGrid[i][j] == 1){
obstacleGrid[i][j] = -1;
}
}
}
for(int i = 0;i < row;i++){
if(obstacleGrid[i][0] == -1) {
break;
} else {
obstacleGrid[i][0] = 1;
}
}
for(int i = 0;i < col;i++){
if(obstacleGrid[0][i] == -1) {
break;
} else {
obstacleGrid[0][i] = 1;
}
}
for(int i = 1;i < row;i++) {
for(int j = 1;j < col;j++) {
if(obstacleGrid[i][j] != -1) {
if(obstacleGrid[i-1][j] ==-1 && obstacleGrid[i][j-1] == -1) {
continue;
}else if(obstacleGrid[i-1][j] ==-1) {
obstacleGrid[i][j] = obstacleGrid[i][j-1];
} else if(obstacleGrid[i][j-1] == -1) {
obstacleGrid[i][j] = obstacleGrid[i-1][j];
} else {
obstacleGrid[i][j] = obstacleGrid[i-1][j] + obstacleGrid[i][j-1];
}
}
}
}
return obstacleGrid[row-1][col-1];
}