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?

我的第一反应就是递归~~~每走一步下面的情况相当于变成m-1,n或者m，n-1的情形

也就是说：ways[m][n]=ways[m-1][n]+ways[m][n-1]+2

为啥是2呢？因为可以向右走也可以向下走两种情况——直到在终点的正上方或者正左方，这个时候只有一种走法

所以我的第一个方法：

int x=0;

int y=0;

public int uniquePaths(int m, int n){

if(x==m-1||y==n-1)	return 1;

return uniquePaths(m-1,n)+uniquePaths(m,n-1)+2;

}

public class UniquedePaths {

public int uniquePaths(int m,int n){

int[][] rec = new int[m][n];

for(int i=0;i<m;i++)

rec[i][0]=1;

for(int j=0;j<n;j++)

rec[0][j]=1;

for(int p=1;p<m;p++)

for(int q=1;q<n;q++)

rec[p][q]=rec[p-1][q]+rec[p][q-1];

return rec[m-1][n-1];

}

}

【Leetcode】Unique Paths in JAVA