Leetcode 63. 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.

public class Main {

public static void main(String[] args) throws Exception {
int[][] input = {{0}
,{0}};
System.out.println(uniquePathsWithObstacles(input));
}

public static int uniquePathsWithObstacles(int[][] obstacleGrid) {
int[][] result = new int[obstacleGrid.length][obstacleGrid[0].length];
boolean flag1 = false;
boolean flag2 = false;
for(int i = 0;i<obstacleGrid.length;i++){						//初始化结果矩阵的第一行，如果之前有障碍以后就都是0了
if(obstacleGrid[i][0]!=1 && !flag1){
result[i][0] = 1;
}else{
flag1 = true;
result[i][0] = 0;
}
}
for(int i = 0;i<obstacleGrid[0].length;i++){					//初始化结果矩阵的第一列，如果之前有障碍以后就都是0了
if(obstacleGrid[0][i]!=1 && !flag2){
result[0][i] = 1;
}else{
flag2 = true;
result[0][i] = 0;
}
}
for(int i = 1;i<obstacleGrid.length;i++){
for(int j = 1;j<obstacleGrid[0].length;j++){
if(obstacleGrid[i][j] == 1){							//如果有障碍则结果矩阵相应位置是0
result[i][j] = 0;
continue;
}
result[i][j] = result[i-1][j]+result[i][j-1];
}
}
return result[obstacleGrid.length-1][obstacleGrid[0].length-1];
}

}