[LeetCode][Java] Unique Paths II
题目:
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.
题意:
紧跟着题目《Unique Paths》,现给出这样一题目:
假设在格子中加入一些障碍,会出现多少存在且唯一的不同路径呢?
障碍和空白格子分别被标记为1
and 0
.
比方一个3x3的格子中的中间存在一个障碍,例如以下所看到的:
[
[0,0,0],
[0,1,0],
[0,0,0]
]
总的路径数为2.
算法分析:
思路与题目《Unique Paths》类似,不同之处为:
初始化边界上行和列时,出现障碍。后面路径数dp的都是0
中间的格子出现障碍时,该格子dp表示的路径数直接填0
AC代码:
public class Solution
{
public int uniquePathsWithObstacles(int[][] obstacleGrid)
{
if(obstacleGrid==null||obstacleGrid.length==0)
return 0;
int m = obstacleGrid.length;
int n = obstacleGrid[0].length;
int [][] dp = new int[m][n];
for(int i = 0; i < m; i++)
{
if(obstacleGrid[i][0]!=1)
dp[i][0] = 1;
else
break;
}
for(int j = 0; j < n; j++)
{
if(obstacleGrid[0][j]!=1)
dp[0][j] = 1;
else
break;
}
for(int i = 1; i < m; i++)
{
for(int j = 1; j< n; j++)
{
if(obstacleGrid[i][j]!=1)
dp[i][j] = dp[i-1][j] + dp[i][j-1];
else
dp[i][j]=0;
}
}
return dp[m-1][n-1];
}
}
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