322. Coin Change

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最新推荐文章于 2024-03-19 21:22:22 发布

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本文链接：https://blog.youkuaiyun.com/fengyuhe1001/article/details/73776005

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本文探讨了CoinChange问题，即给定不同面额的硬币和一个总金额，如何用最少数量的硬币凑出该金额。介绍了两种解法：一种是超时的暴力解法；另一种是使用动态规划进行优化的高效解法。

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322. Coin Change

题目：

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.

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

Example 1:

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

Example 2:

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

代码：

暴力解法（超时）：

class Solution {
public:
    int coinChange(vector<int>& coins, int amount) {
        sort(coins.begin(), coins.end());
        vector<vector<int>> answer;
        vector<int> sol;
        getCom(sol, answer, coins, amount, 0);
        if (answer.size() == 0) return -1;
        else {
            vector<int> min = answer[0];
            for (int i = 0; i < answer.size(); i++) {
                if (answer[i].size() < min.size())
                    min = answer[i];
            }
            return min.size();
        }
    }
    void getCom(vector<int>& sol, vector<vector<int>>& res, vector<int>& coins, int amount, int start) {
        if (!amount) {
            res.push_back(sol);
            return;
        }
        for (int i = start; i < coins.size() && amount >= coins[i]; i++) {
            sol.push_back(coins[i]);
            getCom(sol, res, coins, amount - coins[i], i);
            sol.pop_back();
        }
    }
};

优化版：

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