A robot is located at the top-left corner of a m x n grid (marked 'Start' in the diagram below).

The robot can only move either down or right at any point in time. The robot is trying to reach the bottom-right corner of the grid (marked 'Finish' in the diagram below).

How many possible unique paths are there?
 

Above is a 3 x 7 grid. How many possible unique paths are there?

Note: m and n will be at most 100.

這個題本來是想用回溯法來做，用一個count進行累加，規模小的時候可以，但規模一大就超時了。回溯法程式碼：

class Solution {
public:
    int count; 
    int uniquePaths(int m, int n) {
        count=0;
        find(0,0,m,n);
        return count;
    }
    void
 
 find(int row,int col,int m,int n){
        if(row==m-1 && col==n-1)
            count++;
        if(row<m-1)
            find(row+1,col,m,n);
        if(col<n-1)
            find(row,col+1,m,n);
    }
};

於是轉而用動態規劃進行解題。到達每個節點的路徑的總數都是到達該節點左邊節點的路徑總數，加上到達該節點上面節點的路徑總數。設定一個初始二維陣列，值都設為1，然後依次進行累加即可。

class Solution {
public:
    int uniquePaths(int m, int n) {
        vector<vector<int>> count(m,vector<int>(n,1));
        for(int i=1;i<m;i++)
            for(int j=1;j<n;j++)
                count[i][j]=count[i-1][j]+count[i][j-1];
        return count[m-1][n-1];
    }
};

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.

這個題目是在1的裡面添加了障礙物，方法還是動態規劃，但是初值都賦為0.然後對第一行和第一列進行初始化。在累加的時候，判斷一下當前位置是不是障礙物，如果是，就賦為0，不是再進行累加，程式碼如下：

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