leetcode ---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?

(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;

}

}

 



posted @ 2016-02-24 22:06  wangb021  阅读(196)  评论(0编辑  收藏  举报