leetcode || 63、Unique Paths II

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.

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 Array Dynamic Programming

题意：紧挨着上一题，只不过对矩阵作了一点改变：矩阵的0代表可以通行，1代表不可通行，
求从起点（0,0）到终点（m-1,n-1）的路径总数

thinking：

（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];        }};