Unique Paths II (DP)

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.

//大意是从矩阵的起点到终点共有多少条无重复的路径，1代表障碍点开始见到题目打着动态规划的标签，当时就蒙了(DP一直不懂)；

思路：确定状态：每个位置数据代表从起点到达该点的路径数，障碍点赋为0，

转移方程obstacleGrid[i][j]=obstacleGrid[i-1][j]+obstacleGrid[i][j-1];然后为了避免开始的时候的计算带来的边界的溢出的判断，就先把首行和首列先初始化了；

class Solution {public:    int uniquePathsWithObstacles(vector<vector<int> > &obstacleGrid) {        int m=obstacleGrid.size();        int n=obstacleGrid[0].size();        int i=0,j=0;        if(m==0||obstacleGrid[0][0]==1)            return 0;        if(m==1&&n==1)            return 1;        for(i=0;i<m;i++){            if(obstacleGrid[i][0]==1)                break;            obstacleGrid[i][0]=1;        }        for(;i<m;i++)            obstacleGrid[i][0]=0;//初始化列        for(i=1;i<n;i++){            if(obstacleGrid[0][i]==1)                break;            obstacleGrid[0][i]=1;        }        for(;i<n;i++)            obstacleGrid[0][i]=0;//初始化行        for(i=1;i<m;i++){            for(j=1;j<n;j++){                if(obstacleGrid[i][j]==1){                    obstacleGrid[i][j]=0;                    continue;                }                else{                    obstacleGrid[i][j]=obstacleGrid[i-1][j]+obstacleGrid[i][j-1];                }            }        }        return obstacleGrid.back().back();    }};