leetcode 62. Unique Paths DP动态规划

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.

这道题求解的是从开始到结束点的所有路径的数量，可以使用动态规划了来做，也可以使用递归来做。 这里使用dp[i][j]的值即为路径的数量
转移方程是 dp[i][j] = dp[i-1][j] + dp[i][j-1]
边界条件是 dp[i][1] = 1, dp[1][j] = 1

其实可以发现这是一个斐波那契数列，嗯嗯，就是这么简单，但是包装起来一点都不想斐波那契数列。

代码如下：

public class Solution{    public int uniquePaths(int m, int n)    {        if(m<=0 || n<=0)            return 0;        //return getByRecursion(m, n);        return byDP(m, n);    }    /*     * dp[i][j]的值即为路径的数量      * dp[i][j] = dp[i-1][j] + dp[i][j-1]     * 边界条件：dp[i][1] = 1, dp[1][j] = 1     * */    public int byDP(int m, int n)    {           int [][] mat=new int[m][n];        for(int i=0;i<m;i++)            mat[i][0]=1;        for(int j=0;j<n;j++)            mat[0][j]=1;        for(int i=1;i<m;i++)        {            for(int j=1;j<n;j++)                mat[i][j]=mat[i-1][j]+mat[i][j-1];        }        return mat[m-1][n-1];    }    /*     * 这个问题的递归算法类似斐波拉契数列     * 这个递归结构也就告诉了我们的动态规划的转换方程     * 但是会超时     * */     public int getByRecursion(int m, int n)     {            if(m==1 && n==1)                return 1;            else                 return uniquePaths(m-1, n)+uniquePaths(m, n-1);     }}

下面是C++的做法，就是一个普通的DP

代码如下：

#include <iostream>using namespace std;class Solution {public:    int uniquePaths(int m, int n)     {        int** dp = new int*[m];        for (int i = 0; i < m; i++)            dp[i] = new int[n];        for (int i = 0; i < n; i++)            dp[0][i] = 1;        for (int i = 0; i < m; i++)            dp[i][0] = 1;        for (int i = 1; i < m; i++)        {            for (int j = 1; j < n; j++)            {                dp[i][j] = dp[i - 1][j] + dp[i][j - 1];            }        }        int res = dp[m - 1][n - 1];        for (int i = 0; i < m; i++)            delete[] dp[i];        delete[] dp;        return res;    }};