《leetCode》：Unique Paths II

题目描述

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.

思路

int uniquePathsWithObstacles(int** obstacleGrid, int obstacleGridRowSize, int obstacleGridColSize) {
if(obstacleGrid==NULL||obstacleGridRowSize<1||obstacleGridColSize<1){
return 0;
}
int m=obstacleGridRowSize;
int n=obstacleGridColSize;
//开辟一段空间来保存结果
int **result=(int **)malloc(m*sizeof(int *));
if(result==NULL){
exit(EXIT_FAILURE);
}
for(int i=0;i<m;i++){
result[i]=(int *)malloc(n*sizeof(int));
if(result[i]==NULL){
exit(EXIT_FAILURE);
}
memset(result[i],0,n*sizeof(int));//初始化为零 ,很重要
}
//最后一列为 从下往上故障之前为1，之后为零，表示不可达；
for(int i=m-1;i>=0;i--){
if(obstacleGrid[i][n-1]==0){
result[i][n-1]=1;
}
else{
break;
}
}
//最后一行为 从右往左故障之前为1，之后为零，表示不可达；
for(int i=n-1;i>=0;i--){
if(obstacleGrid[m-1][i]==0){
result[m-1][i]=1;
}
else{
break;
}
}
for(int i=m-2;i>=0;i--){
for(int j=n-2;j>=0;j--){
if(obstacleGrid[i][j]==0){//没有障碍物时，才是右边的点和下边的点的路径之和
result[i][j]=result[i+1][j]+result[i][j+1];
}
//obstacleGrid[i][j]==1时该点的可达路径为零

}
}
return result[0][0];
}