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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.
这是一道非常典型的动态规划的问题
知道递推关系式
paths[i][j] = paths[i - 1][j] + paths[i][j - 1]
注意控制边界条件即可
1 public class Solution { 2 public int uniquePaths(int m, int n) { 3 int paths[][] = new int[m + 1][n + 1]; 4 paths[0][0] = 1; 5 6 for(int i = 1; i < paths.length; i++){ 7 for(int j = 1; j < paths[0].length ; j++){ 8 if(i == 1|| j == 1) 9 paths[i][j] = 1; 10 if(j != 1 && i != 1) 11 paths[i][j] = paths[i - 1][j] + paths[i][j - 1]; 12 }//for 13 }//for 14 15 return paths[m][n]; 16 } 17 }
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原文地址：http://www.cnblogs.com/luckygxf/p/4234720.html
