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?
(二维图像,只能往右或下走一步,从左上角到右下角一共多少种走法?)
方法一:确定到任意一个点能有多少种走法?
当x=0或y=0时,只能有一种;
当x!=0 且 y!=0时,有 matrix[y][x]=matrix[y-1][x] + matrix[y][x-1]种走法;用 (上+左)
代码:
public class Solution {
int[][] matrix;
public int uniquePaths(int m, int n) {
matrix = new int[n][m];
for(int y=0; y<n; y++) {
for(int x=0; x<m; x++) {
//Fill top row and left most row with 1s
if(x == 0 || y==0) matrix[y][x]=1;
else {
matrix[y][x]=matrix[y-1][x] + matrix[y][x-1]; //相当于把每个点的走法种类都表示出来
}
}
}
return matrix[n-1][m-1]; //返回的时候要注意,不要越界
}
}
方法二:知道从左上到右下,一共有m+n-2步,就分为两类,一类是右(m-1)步;一类是下(n-1)步; 题目转换为:在m+n-2步中,挑选哪些步是右(下)的
代码:
public class Solution {
public int uniquePaths(int m, int n) {
long result = 1;
for(int i=0;i<Math.min(m-1,n-1);i++)
result = result*(m+n-2-i)/(i+1); //公式
return (int)result;
}
}