【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?

Above is a 3 x 7 grid. How many possible unique paths are there?

Note: m and n will be at most 100.

code: // 一个简单的动态规划问题。

 

class Solution {

public:

    int uniquePaths(int m, int n) {

        // Note: The Solution object is instantiated only once and is reused by each test case.

        if(m == 0 || n == 0)

            return 0;

        int **dp = new int*[m];

        for(int i = 0; i < m; i++)

        {

            dp[i] = new int[n];

        }

        dp[0][0] = 1;

        for(int i = 1; i < m; i++)

            dp[i][0] = 1;

        for(int j = 1; j < n; j++)

            dp[0][j] = 1;

        for(int i = 1; i < m; i++)

        {

            for(int j = 1; j < n; j++)

            {

                dp[i][j] = dp[i][j-1] + dp[i-1][j];

            }

        }

        int res = dp[m-1][n-1];

        for(int i = 0; i < m; i++)

        {

            delete [] dp[i];

        }

        delete [] dp;

        return res;

    }

};

来一个java版本的，不用显式回收从堆中申请的内存，由java自己的垃圾回收机制来回收，只需讲引用变量和引用对象之间切断，让该内存成为垃圾。

 

public class Solution {

    public int uniquePaths(int m, int n) {

        // Note: The Solution object is instantiated only once and is reused by each test case.

        if(m == 0 || n == 0)

            return 0;

        int[][] dp = new int[m][n];

        dp[0][0] = 1;

        for(int i = 1; i < m; i++)

            dp[i][0] = 1;

        for(int j = 1; j < n; j++)

            dp[0][j] = 1;

        for(int i = 1; i < m; i++)

        {

            for(int j = 1; j < n; j++)

            {

                dp[i][j] = dp[i][j-1] + dp[i-1][j];

            }

        }

        int res = dp[m-1][n-1];

        for(int i = 0; i < m; i++)

        {

             dp[i] = null;

        }

        dp = null;

        return res;

    }

    

}