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Problem Description:
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?
Code:
public class Solution { /** * @param n, m: positive integer (1 <= n ,m <= 100) * @return an integer */ public int uniquePaths(int m, int n) { // write your code here if (m == 0 || n == 0) { return 0; } int[][] sum = new int[m][n]; for (int i = 0; i < m; i++) { sum[i][0] = 1; } for (int j = 0; j < n; j++) { sum[0][j] = 1; } for (int i = 1; i < m; i++) { for (int j = 1; j < n; j++) { sum[i][j] = sum[i - 1][j] + sum[i][j - 1]; } } return sum[m - 1][n - 1]; } }
Hint:
Here we use dynamic programming, the most important thing is know that sum[i][j] = sum[i - 1][j] + sum[i][j -1], because you can go right from position(i, j - 1) or go down from position(i - 1, j) to get to position(i, j).
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原文地址：http://www.cnblogs.com/dingjunnan/p/5496058.html
