unique-paths with dynamic programming

gistfile1.txt

https://oj.leetcode.com/problems/unique-paths/

/*
let's count the number of cases:
m,n=00/10/01 .. 1
m,n=1,1 1
m,n=1,2 1
m,n=1,3 1
m,n=2,1 1
m,n=2,2 (2,1) + (1,2) = 2
m,n=2,3 (2,2) + (1,3) = 3
m,n=2,4 (2,3) + (1,4) = 4
m,n=3,2 (2,2) + (3,1) = 5
m,n=3,3 (3,2) + (2,3) = 6
m,n=3,4 (3,3) + (2,4) = 6+4 = 10

...
so, if mn == ij then mn = (i,j-1) + (i-1,j).
go recursive solution with loops, not function call because of memory/speed efficiency

O(mn) solution : using 2 forloops to calculate map[m][n] from 1,1 to m,n   
*/

class Solution {
public:
    int uniquePaths(int m, int n) {
        if(m <= 1 || n <= 1) {
            return 1; 
        }
        
        vector<int> col(n,1);
        vector<vector<int>> map;
        for(int i=0; i<m ; ++i) {
            map.push_back(col);
        }
        
        for(int i=1 ; i<m; i++) {
           for(int j=1 ; j<n; ++j) {
               map[i][j] = map[i-1][j]+map[i][j-1];
           }
        }
        
        return map[m-1][n-1];
    }
};