322. Coin Change

最新推荐文章于 2025-09-10 15:16:07 发布

转载最新推荐文章于 2025-09-10 15:16:07 发布 · 92 阅读

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原文链接：http://www.cnblogs.com/ruruozhenhao/p/10389187.html

文章标签：

#c/c++

本文探讨了硬币找零问题的两种解决策略：动态规划与深度优先搜索结合贪婪算法及剪枝优化。通过两个示例，分别展示了如何使用最少数量的硬币组合达到指定金额，以及在无法组合的情况下返回-1。文章提供了详细的C++代码实现，并分析了每种方法的时间复杂度。

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:

Input: coins = [1, 2, 5], amount = 11
Output: 3 
Explanation: 11 = 5 + 5 + 1

Example 2:

Input: coins = [2], amount = 3
Output: -1

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

Approach #1: DP. [C++]

    int coinChange1(vector<int>& coins, int amount) {
        vector<int> dp(amount+1, amount+1);
        dp[0] = 0;
        
        for (int coin : coins) {
            for (int i = coin; i <= amount; ++i) {
                dp[i] = min(dp[i], dp[i-coin]+1);
            }
        }
        
        return dp[amount] == amount+1 ? -1 : dp[amount];
    }

Approach #2: DFS + Greedy + Pruning. [C++]

class Solution {
public:
    int coinChange(vector<int>& coins, int amount) {
        sort(coins.rbegin(), coins.rend());
        int ans = INT_MAX;
        
        coinChange(coins, 0, amount, 0, ans);
        
        return ans == INT_MAX ? -1 : ans;
    }
    
private:
    void coinChange(const vector<int>& coins, 
                    int idx, int amount, int count, int& ans) {
        const int coin = coins[idx];
        if (idx == coins.size()-1) {
            if (amount % coin == 0) 
                ans = min(ans, count + amount / coin);
        } else {
            for (int k = amount / coin; k >= 0 && count+k < ans; --k) {
                coinChange(coins, idx+1, amount-k*coin, count+k, ans);
            }
        }
        
    }
    
};

Analysis:

http://zxi.mytechroad.com/blog/dynamic-programming/leetcode-322-coin-change/

转载于:https://www.cnblogs.com/ruruozhenhao/p/10389187.html