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Follow up for "Unique Paths":紧接着上一题“唯一路劲”，现在考虑有一些障碍在网格中，无法到达，请重新计算到达目的地的路线数目
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.
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//思路首先：
//此题与原问题相较，变得是什么？
//1，此障碍物下面和右边将不在获得来自于此的数量,也可以理解为贡献为0
//2，有障碍的地方也将无法到达（这一条开始时没想到，总感觉leetcode题目意思不愿意说得明了），可能输数直接为0
class Solution {
public:
int uniquePathsWithObstacles(vector<vector<int>>& obstacleGrid) {
int m=obstacleGrid.size();
int n=obstacleGrid[0].size();
vector< vector<int> > result(m+1);
for(int i=0;i <=m ;i++)
result[i].resize(n+1);//设置数组的大小m+1行，n+1列
//初始化一定要正确，否则错无赦
result[1][1]= obstacleGrid[0][0]==1? 0:1;
for(int i=2;i<=n;i++)
result[1][i]=obstacleGrid[0][i-1]==1?0:result[1][i-1];//由上一次来推到
for(int i=2;i<=m;i++)
result[i][1]=obstacleGrid[i-1][0]==1?0:result[i-1][1];
for(int i=2;i<=m;i++)
for(int j=2;j<=n;j++)
result[i][j]=obstacleGrid[i-1][j-1]==1?0:result[i-1][j]+result[i][j-1]; //一旦当前有石头就无法到达，直接置零
return result[m][n];
}
};
<LeetCode OJ> 63. Unique Paths II
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原文地址：http://blog.csdn.net/ebowtang/article/details/50485468
