Leetcode63: Unique Paths II

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：

接着leetcode62题，还是在问从左上角到右下角的路径数目，不过这一题提出，如果一个格子的值为1的话，就说明这个格子有障碍，不可以到达和穿过的。

分析：

这一题还是同leetcode62一样使用DP来做，特别的地方是对矩阵中1，即障碍的处理。

代码：

class Solution {public:    int uniquePathsWithObstacles(vector<vector<int>>& obstacleGrid) {        int m=obstacleGrid.size();        if(m==0) return 0;        int n=obstacleGrid[0].size();        if(n==0) return 0;                int a[110][110]={0};        if(obstacleGrid[0][0]==1)a[0][0]=0;        else a[0][0]=1;                for(int i=1; i<m; i++){            if(obstacleGrid[i][0]==1) a[i][0]=0;            else a[i][0]=a[i-1][0];        }        for(int i=1; i<n; i++){            if(obstacleGrid[0][i]==1) a[0][i]=0;            else a[0][i]=a[0][i-1];        }               for(int i=1; i<m; i++){            for(int j=1; j<n; j++){                if(obstacleGrid[i][j]==1) a[i][j]=0;                else{                    if(obstacleGrid[i-1][j]!=1 && obstacleGrid[i][j-1]!=1)                    a[i][j]=a[i-1][j]+a[i][j-1];                    else if(obstacleGrid[i-1][j]==1 && obstacleGrid[i][j-1]!=1)                    a[i][j]=a[i][j-1];                    else if(obstacleGrid[i-1][j]!=1 &&obstacleGrid[i][j-1]==1)                    a[i][j]=a[i-1][j];                    else                    a[i][j]=0;                }            }        }                return a[m-1][n-1];    }};