leetcode oj java Unique Paths

来源：互联网发布：tensorflow viterbi 编辑：程序博客网时间：2024/05/16 19:09

一、问题描述：

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.

二、解决思路：

1很容易想到递归： f(m,n) = f(m-1,n) + f(m,n-1) ，终止条件是m==1 || n==1但是time limit

2动态规划，：用二维数组存放路径数目， a[m][n] = a[m-1][n] + a[m][n-1]

三、代码：

package T12;/** * @author 作者 : xcy * @version 创建时间：2016年12月20日 下午3:42:16 *          类说明 */public class t62 {    public static int re = 0;    public static void main(String[] args) {        // TODO Auto-generated method stub        System.out.println(uniquePaths(10, 10));    }    //    public static int uniquePaths(int m, int n) {    //        if (m == 1 || n == 1) {    //            return 1;    //        }    //        return uniquePaths(m - 1, n) + uniquePaths(m, n - 1);    //    }    public static int uniquePaths(int m, int n) {        int[][] re = new int[m][n];        for (int i = 0; i < m; i++) {            re[i][0] = 1;        }        for (int j = 0; j < n; j++) {            re[0][j] = 1;        }        for (int i = 1; i < m; i++) {            for (int j = 1; j < n; j++) {                re[i][j] = re[i - 1][j] + re[i][j - 1];            }        }        return re[m - 1][n - 1];    }}

0 0