LeetCode 61 Unique Paths

最新推荐文章于 2021-02-24 07:42:20 发布

mlweixiao

最新推荐文章于 2021-02-24 07:42:20 发布

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分类专栏： leetcode LeetCode专栏文章标签： java leetcode

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本文介绍了一种计算机器人从网格左上角到右下角唯一路径数量的方法。提供了两种实现思路，一种是利用Java的BigInteger类进行计算，另一种是通过动态规划的方法填充二维数组来解决问题。

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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?

Note: m and n will be at most 100.

思路1：使用Java 的BigInteger类

import java.math.BigInteger;

public class Solution {
   public int uniquePaths(int m, int n) {
       BigInteger  num=BigInteger.ONE;
       int big=m>n?m:n;
       int small=m<n?m:n;
       for(int i=big;i<m+n-1;i++){
    	   num=num.multiply(BigInteger.valueOf(i));
       }
       
       for(int i=2;i<small;i++){
    	   num=num.divide(BigInteger.valueOf(i));
       }
       return num.intValue();
    }
}

思路2：利用如下公式

	public int uniquePaths_2(int m, int n) {
		int[][] grid = new int[m][n];
		int i, j;

		for (i = 0; i < grid.length; i++) {
			grid[i][0] = 1;
		}
		for (i = 1; i < grid[0].length; i++) {
			grid[0][i] = 1;
		}

		for (i = 1; i < grid.length; i++) {
			for (j = 1; j < grid[0].length; j++) {
				grid[i][j] = grid[i - 1][j] + grid[i][j - 1];
			}
		}
		return grid[m - 1][n - 1];
	}