LeetCode_62---Unique Paths

来源：互联网发布：易企秀相似软件编辑：程序博客网时间：2024/05/20 18:47

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.

Hide Tags

Array Dynamic Programming

翻译：

Code:

//224msAC---http://www.cnblogs.com/chkkch/archive/2012/11/15/2772263.htmlpublic static int uniquePaths1(int m, int n) {if (m <= 1 || n <= 1) {return 1;}int temp[][] = new int[m][n];for (int i = 0; i < n; i++) {temp[0][i] = 1;}for (int i = 1; i < m; i++) {temp[i][0] = 1;}for (int i = 1; i < m; i++) {for (int j = 1; j < n; j++) {temp[i][j] = temp[i - 1][j] + temp[i][j - 1];//类似于Fib算法及其演变问题，如：一次只能跳一到两步，问多少次可以跳到n;}}return temp[m - 1][n - 1];}

/** *  */package From61;/** * @author MohnSnow * @time 2015年6月29日 上午9:08:45 *  */public class LeetCode62 {/** * @param argsmengdx *            -fnst *///244msaAC---解决方法是利用排列组合的原理去解决//C(m+n-2,m-1)总共需要走m+n-2步，选取其中的m-1步public static int uniquePaths(int m, int n) {if (m <= 1 || n <= 1) {return 1;}long sum_up = 1;long sum_down = 1;int sum = m + n - 2;int min = (m > n ? n : m) - 1;for (int i = sum; i > sum - min; i--) {System.out.println("i: " + i);sum_up *= i;}System.out.println("sum_up: " + sum_up);for (int i = 1; i <= min; i++) {System.out.println("i: " + i);sum_down *= i;}System.out.println("sum_down: " + sum_down);return (int) (sum_up / sum_down);}public static void main(String[] args) {System.out.println("uniquePaths: " + uniquePaths(10, 10));}}

0 0