leetcode 63動態規劃障礙路徑
與無障礙類似,需多寫的是根據某點狀態判斷到達某點的路徑數。
式為 w[i][j]=(1-obstacleGrid[i][j])*(w[i][j-1]+w[i-1][j])
class Solution {
public int uniquePathsWithObstacles(int[][] obstacleGrid) {
int row = obstacleGrid.length;
if(row==0){
return 0;
}
int col = obstacleGrid[0].length;
int[][] w = new int[row][col];
w[0][0]=1-obstacleGrid[0][0];
for(int i=1;i<col;i++){
w[0][i] = (1-obstacleGrid[0][i])*w[0][i-1];
}
for(int i=1;i<row;i++){
w[i][0] = (1-obstacleGrid[i][0])*w[i-1][0];
}
for(int i=1;i<row;i++){
for(int j=1;j<col;j++){
w[i][j]=(1-obstacleGrid[i][j])*(w[i][j-1]+w[i-1][j]);
}
}
return w[row-1][col-1];
}
}