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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.
思路: 计从位置(i,j)到位置(m-1,n-1)的有f(i,j)种方法,则f(i,j)=f(i,j+1)+f(i+1,j),其中,0<=i<=m-1;0<=j<=n-1,而且f(m-1,j)=1,f(i,n-1)=1,所以,代码如下:
public class Solution { public int uniquePaths(int m, int n) { int[][] a = new int[m][n]; for(int i=0;i<m;i++){ a[i][n-1]=1; } for(int i=0;i<n;i++){ a[m-1][i]=1; } for(int i=m-2;i>=0;i--){ for(int j=n-2;j>=0;j--){ a[i][j]=a[i+1][j]+a[i][j+1]; } } return a[0][0]; } }
public class Solution { public int uniquePaths(int m, int n) { int[] a=new int[m]; Arrays.fill(a,1); for(int i=n-2;i>=0;i--){ int b=1; for(int j=1;j<=m-1;j++){ a[j]=a[j]+b; b=a[j]; } } return a[m-1]; } }
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原文地址:http://blog.csdn.net/u010786672/article/details/44133057