【LeetCode】Unique Paths

来源:互联网 发布:索尼g8145支持4g网络吗 编辑:程序博客网 时间:2024/06/01 07:38

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;    }    }



原创粉丝点击