zoukankan      html  css  js  c++  java
  • [CareerCup] 9.8 Represent N Cents 美分的组成

    9.8 Given an infinite number of quarters (25 cents), dimes (10 cents), nickels (5 cents) and pennies (1 cent), write code to calculate the number of ways of representing n cents.

    这道题给定一个钱数,让我们求用quarter,dime,nickle和penny来表示的方法总和,很明显还是要用递归来做。比如我们有50美分,那么

    makeChange(50) =

      makeChange(50 using 0 quarter) +

      makeChange(50 using 1 quarter) +

      makeChange(50 using 2 quarters)

    而其中第一个makeChange(50 using 0 quarter)又可以拆分为:

    makeChange(50 using 0 quarter) =

      makeChange(50 using 0 quarter, 0 dimes) + 

      makeChange(50 using 0 quarter, 1 dimes) + 

      makeChange(50 using 0 quarter, 2 dimes) + 

      makeChange(50 using 0 quarter, 3 dimes) + 

      makeChange(50 using 0 quarter, 4 dimes) + 

      makeChange(50 using 0 quarter, 5 dimes)

    而这里面的每项又可以继续往下拆成nickle和penny,整体是一个树形结构,计算顺序是从最底层开始,也就是给定的钱数都是由penny组成的情况慢慢往回递归,加一个nickle,加两个nickle,再到加dime和quarter,参见代码如下:

    解法一:.

    class Solution {
    public:
        int makeChange(int n) {
            vector<int> denoms = {25, 10, 5, 1};
            return makeChange(n, denoms, 0);
        }
        int makeChange(int amount, vector<int> denoms, int idx) {
            if (idx >= denoms.size() - 1) return 1;
            int val = denoms[idx], res = 0;
            for (int i = 0; i * val <= amount; ++i) {
                int rem = amount - i * val;
                res += makeChange(rem, denoms, idx + 1);
            }
            return res;
        }
    };

    上述代码虽然正确但是效率一般,因为存在大量的重复计算,我们可以用哈希表来保存计算过程中的结果,下次遇到相同结果时,直接从哈希表中取出来即可,参见代码如下:

    解法二:

    class Solution {
    public:
        int makeChange(int n) {
            vector<int> denoms = {25, 10, 5, 1};
            vector<vector<int> > m(n + 1, vector<int>(denoms.size()));
            return makeChange(n, denoms, 0, m);
        }
        int makeChange(int amount, vector<int> denoms, int idx, vector<vector<int> > &m) {
            if (m[amount][idx] > 0) return m[amount][idx];
            if (idx >= denoms.size() - 1) return 1;
            int val = denoms[idx], res = 0;
            for (int i = 0; i * val <= amount; ++i) {
                int rem = amount - i * val;
                res += makeChange(rem, denoms, idx + 1, m);
            }
            m[amount][idx] = res;
            return res;
        }
    };
  • 相关阅读:
    unigui在阿里云服务器上部署
    nativeexcel将excel导入数据集
    nativeexcel数据集导出excel
    超市收银系统
    cxgrid回车移到下一个单元格
    cxGrid单元格获得输入焦点
    TDiocpCoderTcpServer返回数据记录有条数限制的问题
    Ubuntu -- 配置Nginx和https及frp
    REMOTE HOST IDENTIFICATION HAS CHANGED 问题解决
    iOS -- iOS11新特性,如何适配iOS11
  • 原文地址:https://www.cnblogs.com/grandyang/p/4840713.html
Copyright © 2011-2022 走看看