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 };