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Unique Paths

时间:2015-03-08 15:45:00      阅读:136      评论:0      收藏:0      [点我收藏+]

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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?

技术分享

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];
    }
}

节省内存:

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];
    }
}




Unique Paths

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原文地址:http://blog.csdn.net/u010786672/article/details/44133057

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