贪心算法

找零钱问题

the problem of “Making Change”.

Coins available are:

dollars (100 cents)
quarters (25 cents)
dimes (10 cents)
nickels (5 cents)
pennies (1 cent)

problem： Making a change of a given amount using the smallest possible number of coins.

#include <iostream>#include <array>#include <vector>using namespace std;int main() {    // your code goes here    array<int, 5> coins = {100, 25, 10, 5, 1};    int n;    cin >> n;    int sum = 0;    vector<int> changes;    for (int i = 0; i < coins.size();) {        if (sum + coins[i] <= n) {            sum += coins[i];            changes.push_back(coins[i]);            i = 0;        } else {            i++;        }    }    int i = 0;    while (sum != n) {        if (sum + coins[i] <= n) {            sum += coins[i];            changes.push_back(coins[i]);            i = 0;        } else {            i++;        }    }    cout << "need smallest " << changes.size() << " coins: ";    for (auto c : changes) {        cout << c << " ";    }    cout << endl;    return 0;}

贪心算法的可行性定义

使用贪心算法求得的解不仅是问题解集合中的一种解，也是该问题的最优解。

解决动态规划问题由底向上，而贪心法通常由顶向下，且每一个问题都做出最贪心的选择，并逐步减小每一个问题到更小规模。

参考

Greedy Introduction