Java for LeetCode 062 Unique Paths

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?

解题思路一：

很明显，答案是C(m+n-2,n-1)，是一个排列组合问题，直接使用int进行操作的话，会发生数据溢出，不过好在我们有BigInteger，JAVA实现如下：

import java.math.BigInteger;

public class Solution {
    public int uniquePaths(int m, int n) {
	        return Cmn(m,n).intValue();
	    }
	    static BigInteger Cmn(int m, int n){
	    	return Amn(m + n - 2, n - 1).divide(factorial(n - 1));
	    }
	    static BigInteger Amn(int m, int n) {
	        if (n == 0)
	            return BigInteger.valueOf(1);
	        if (n == 1)
	            return BigInteger.valueOf(m);
	        else
	            return Amn(m - 1, n - 1).multiply(BigInteger.valueOf(m));
	    }
	 
	    static BigInteger factorial(int n) {
	        if (n == 1 || n == 0)
	            return BigInteger.valueOf(1);
	        else
	            return factorial(n - 1).multiply(BigInteger.valueOf(n));
	    }
}

 

 解题思路二：

很明显uniquePaths(int m, int n)=uniquePaths(int m-1, int n)+uniquePaths(int m, int n-1)，结果提交发现Time Limit Exceeded，看来不能直接用递归计算。不过基于上述结论，我们可以采用dp的方法，开一个数组v[j]表示经过i步右移和j步下移后达到(i,j)的路径数目，前进方法为v[j] += v[j - 1]，表示v[j]由从左边过来的v[j-1]和从上面下来的前一个v[j]组成，JAVA实现如下：

	static public int uniquePaths(int m, int n) {
		int[] v = new int[n];
		for (int i = 0; i < v.length; i++)
			v[i] = 1;
		for (int i = 1; i < m; ++i)
			for (int j = 1; j < n; j++)
				v[j] += v[j - 1];
		return v[n - 1];
	}