Unique Paths

A robot is located at the top-left corner of a m x n grid.
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?

Solution:
1. Use formula C(x,t) = t!/(x!*(t-x)!) (x should be large for calculation).
2. Dynamic programming. UP(i,j) = UP(i-1,j) + UP(i,j-1).

 1 class Solution {
 2 public:
 3     int uniquePaths(int m, int n) {
 4         int dp[m][n];
 5         for(int i = 0; i < m; i++)
 6             dp[i][0] = 1;
 7         for(int j = 0; j < n; j++)
 8             dp[0][j] = 1;
 9         for(int i = 1; i < m; i++) {
10             for(int j = 1; j < n; j++) {
11                 dp[i][j] = dp[i-1][j] + dp[i][j-1];
12             }
13         }
14         return dp[m-1][n-1];
15     }
16 };

Unique Paths,布布扣,bubuko.com

Unique Paths

原文：http://www.cnblogs.com/zhengjiankang/p/3650941.html