Unique Paths 算法详解

Unique Paths I
算法题目：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.

大致意思：如上图所示，robot从左上角走到右下角有多少种走法？

思路：F(i,j)=F(i-1,j)+F(i,j-1)

    int uniquePaths(int m, int n) {        if(m<=0||n<=0)return 0;        vector<vector<int>> ret(m,vector<int>(n,1));        for(int i=1;i<m;i++)        {            for(int j=1;j<n;j++)            {                ret[i][j]=ret[i][j-1]+ret[i-1][j];            }        }        return ret[m-1][n-1];    }

Unique Paths II
算法题目：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.

Note: m and n will be at most 100.

思路：和I思路一样，若为障碍，则相应设置F[i][j]=0;

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