unique-paths I &II 路径数，动态规划

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.

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

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 as1and0respectively 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 is2.

Note: m and n will be at most 100.

有障碍物的地方不能走，循环内加判断

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