leetCode 62.Unique Paths （唯一路径） 解题思路和方法

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.

思路：这题首先想到的是递归，但是递归效率太慢，超时不解释。于是换动态规划，也算典型的动态规划的题了。

具体代码如下：

public class Solution {
public int uniquePaths(int m, int n) {

//本题解法为动态规划
//状态转移方程f[i][j] = f[i-1][j] + f[i][j-1];
//f[i][j]的值即为路径的数量

int[][] f = new int[m]
;
for(int i = 0; i < m ; i++)//第一列赋值为1
f[i][0] = 1;
for(int i = 0; i < n; i++)//第一行赋值为1
f[0][i] = 1;

for(int i = 1; i < m ; i++)
for(int j = 1; j < n ; j++){
f[i][j] = f[i-1][j] + f[i][j-1];
//System.out.println("f[" + i +"][" + j+ "]" + f[i][j]);
}
return f[m-1][n-1];//返回结果值
}
}