62. Unique Paths(排列组合法&动态规划法)

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.

本题要求的是机器人在m*n的图中走从左上角到右下角所以可能路径的数量

解法一（组合）

对于m*n的图，机器人走到目标点需要(m+n-2)步，其中（m-1）步向右走，（n-1）步向下走。因此可能路径总数即为从(m+n-2)项中取（m-1）项的方法总数，即(m+n-2)!/(m-1)!/(n-1)!

代码如下

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);         return (int)result;    }}

解法二（动态规划）

使用递归，机器人从坐标(m,n)出发的可能路径为从(m-1,n)出发和从(m,n-1)出发的路径和

代码如下：

public class Solution {    public int[][] dp = new int[101][101];    public int uniquePaths(int m, int n) {        if (m == 0 || n == 0)  return 0;        if (m == 1 || n == 1)  return 1;        if(dp[m][n] != 0) return dp[m][n];        return dp[m][n] = uniquePaths(m-1, n) + uniquePaths(m, n-1);    }}

也可以不使用递归改用循环迭代

public class Solution {    public int uniquePaths(int m, int n) {        int[][] tab = new int[m][n];        for(int i=0; i<m; i++){            for(int j=0; j<n; j++){                if(i==0 || j==0){                    tab[i][j] = 1;                }                else{                    tab[i][j] = tab[i-1][j]+tab[i][j-1];                }            }        }        return tab[m-1][n-1];    }}