十一周- Dynamic Programming- Coin Change

最新推荐文章于 2019-01-02 10:57:46 发布

原创最新推荐文章于 2019-01-02 10:57:46 发布 · 662 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#函数 #一周

算法专栏收录该内容

24 篇文章

订阅专栏

本文探讨了硬币找零问题的解决方法，通过动态规划算法找出组成特定金额所需的最少硬币数量。假设硬币数量无限，文章提供了一个C++实现的例子。

You are given coins of different denominations and a total amount of money amount. Write a function to compute the fewest number of coins that you need to make up that amount. If that amount of money cannot be made up by any combination of the coins, return -1.

Example 1:
coins = [1, 2, 5], amount = 11
return 3 (11 = 5 + 5 + 1)

Example 2:
coins = [2], amount = 3
return -1.

Note:
You may assume that you have an infinite number of each kind of coin.

思路

给定一些不同价值的硬币，假设每个硬币的数量无限，给定一个特定值，需要求是否能用这些硬币组合得到该值，如果能得到给出使用最少的硬币的数量。
使用动态规划，从1到给定值amount，依次求得使用硬币能得到该值的最少硬币数。

源码

class Solution {  
public:  
    int coinChange(vector<int>& coins, int amount) {  
        int a[10000];  
        int i,j,t;  
        for(i = 1;i <= amount;i ++)  
            a[i] = -1;  
        a[0] = 0;  
        for(i = 1;i <= amount;i ++){  
            for(j = 0;j < coins.size();j ++){  
                if((i - coins[j]) >= 0 && a[i - coins[j]] >= 0){  
                    t = a[i - coins[j]] + 1;  
                    if(a[i] == -1 || a[i] > t)  
                      a[i] = t;  
                }  
            }   
        }  
        return a[amount];  
    }  
};