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.
SOLUTION 1:
使用DP解决之。很简单。某一个cell有2种可能到达，从上面来，以及从左边来，只需要把每一个cell的可能数计算好，并且把它们相加即可。
另外，在第一行，第一列，以及左上角需要特别处理，因为我们上面、左边并没有cells.
1 public class Solution { 2 public int uniquePaths(int m, int n) { 3 if (m <= 0 || n <= 0) { 4 return 0; 5 } 6 7 int[][] D = new int[m][n]; 8 9 for (int i = 0; i < m; i++) { 10 for (int j = 0; j < n; j++) { 11 if (i == 0 && j == 0) { 12 D[i][j] = 1; 13 } else if (i == 0) { 14 D[i][j] = D[i][j - 1]; 15 } else if (j == 0) { 16 D[i][j] = D[i - 1][j]; 17 } else { 18 D[i][j] = D[i - 1][j] + D[i][j - 1]; 19 } 20 } 21 } 22 23 return D[m - 1][n - 1]; 24 } 25 }
原文：http://www.cnblogs.com/yuzhangcmu/p/4401242.html
