《leetCode》:Unique Paths II
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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
思路与上篇博文基本一样。
还是用空间换取时间的思路来做,
只不过这里对障碍物的处理方法为:对该点的到终点的可达路径设为零
实现代码如下:
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];}
AC结果如下:
1 0
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