unique paths ii java_[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：
如果在格子中添加一些障碍，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1and0.
比如一个3x3的格子中的中间存在一个障碍，如下所示：
[
[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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原文：http://blog.csdn.net/evan123mg/article/details/46923225