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  • 【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;
        }
        
    }


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  • 原文地址:https://www.cnblogs.com/keanuyaoo/p/3367634.html
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