題目：

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.

題意：給出一個m*n的網格，有一個機器人，每一步只能向右或者向下走，求能從最左上角的格子到最右下角的格子有多少種走法。這道題是典型的動態規劃，將網格看成一個二維數組A[m][n],那麽我們要求的就是從A[0][0]到A[m-1][n-1]的路徑，對於其中任意一點A[i][j](i<m,j<n)來說，到該點的方法有從A[i-1][j]向下走一步或者A[i][j-1]向右走一步，所以d(A[i][j])=d(A[i-1][j]）+d(A[i][j-1])。對每個點遍歷求解即可。這裏需要註意對第一行和第一列上的所有點都只有一種走法。

代碼：

public class Solution {

	public int uniquePaths(int m, int n) {
		if(m==0||n==0) return 0;
		if(m==1||n==1)	//只有一行或者只有一列 只有一種走法
			return 1;
		int[][] A=new int[m][n];		//用戶記錄起點到當前點走法
		
		for(int i=0;i<m;i++){
			for(int j=0;j<n;j++){
				if(i==0||j==0)
					A[i][j]=1;
				else A[i][j]=A[i-1][j]+A[i][j-1];
			}
		}
		return A[m-1][n-1];
	}
}

LeetCode 62. Unique Paths Java