322. Coin Change

最新推荐文章于 2024-03-19 21:22:22 发布

转载最新推荐文章于 2024-03-19 21:22:22 发布 · 44 阅读

0 ·

CC 4.0 BY-SA版权

原文链接：http://www.cnblogs.com/tekkaman-blade/p/7373412.html

本文提出了一种解决硬币找零问题的方法，使用贪心算法替代动态规划，通过将硬币按从大到小排序并尝试使用面额最大的硬币，直至总额达到目标金额。该算法不仅效率高，而且能够找到最优解。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Actually, this problem could use greedy algorithm instead of dynamic programming and greedy algorithm is much faster. The basic idea is sort the coins array from the largest to the smallest and then try coin with the largest value until the total value is larger than "amount". If the total value cannot be equal to "amount" and then go back to the upper layer and try the second largest one... So the optimal answer always comes first.

class Solution {
public:
     int coinChange(vector<int>& coins, int amount) {
        sort(coins.begin(), coins.end(), greater<int>());

        int res = amount+1;
        ccHelper(amount, coins, 0, 0, res);
        return res==amount+1? -1: res;
    }
    void ccHelper(int amount, vector<int>&coins, int start, int step, int& res) {
        if (step+ceil(amount*1.0/coins[start])>=res) return;
        if (amount==0) {res=step; return;}
        for (int i=start; i<coins.size(); i++) {
            if (coins[i]<=amount) ccHelper(amount-coins[i], coins, i, step+1, res);
        }
    }
};

"greater" on the 4th line is an interesting and useful class. Please pay attention to it.

转载于:https://www.cnblogs.com/tekkaman-blade/p/7373412.html