标签:blog io ar os sp java for div on
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?
因为只能向下或者向右走,所依某一点的值就是上面方格的值 和左边的值 相加
dynamic programming.
public class Solution {
public int uniquePaths(int m, int n) {
int[][] grid=new int[100][100];
for(int i=0;i<m;i++){
grid[i][0]=1;
}
for(int j=0;j<n;j++){
grid[0][j]=1;
}
for(int i=1;i<m;i++){
for(int j=1;j<n;j++){
grid[i][j]=grid[i-1][j]+grid[i][j-1];
}
}
return grid[m-1][n-1];
}
}
标签:blog io ar os sp java for div on
原文地址:http://www.cnblogs.com/lilyfindjobs/p/4103068.html