problem：
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.
Array Dynamic Programmingthinking：
（1）blog.csdn.net/hustyangju/article/details/44829339 讨论了只能使用DP法
（2）加了条件限制，则DP算法也要修改：
1、边界条件要改变，一旦出现1，则随后的边界条件全部为0
2、矩阵中间出现1，则该位置的路径数置为0
code：
class Solution {
public:
int uniquePathsWithObstacles(vector<vector<int> > &obstacleGrid) {
vector<vector<int> >::const_iterator con_it=obstacleGrid.begin();
int m=obstacleGrid.size();
int n=(*con_it).size();
vector<int> tmp(n,0);
vector<vector<int> > a(m,tmp);
bool flag=true;
for(int i=0;i<m;i++) //边界条件
{
if(obstacleGrid[i][0]==0 && flag)
a[i][0]=1;
else
{
a[i][0]=0;
flag=false;
}
}
flag=true;
for(int j=0;j<n;j++) //边界条件
{
if(obstacleGrid[0][j]==0 && flag)
a[0][j]=1;
else
{
a[0][j]=0;
flag=false;
}
}
for(int i = 1; i < m; i++)
for(int j = 1; j < n; j++)
{
if(obstacleGrid[i][j]==1) //出现障碍物，置0
a[i][j]=0;
else
a[i][j] = a[i-1][j] + a[i][j-1];
}
return a[m-1][n-1];
}
};leetcode || 63、Unique Paths II
原文地址：http://blog.csdn.net/hustyangju/article/details/44832173
