README.md

322. 零钱兑换

English Version

题目描述

给定不同面额的硬币 coins 和一个总金额 amount。编写一个函数来计算可以凑成总金额所需的最少的硬币个数。如果没有任何一种硬币组合能组成总金额，返回 -1。

你可以认为每种硬币的数量是无限的。

 

示例 1：

输入：coins = [1, 2, 5], amount = 11
输出：3
解释：11 = 5 + 5 + 1

示例 2：

输入：coins = [2], amount = 3
输出：-1

示例 3：

输入：coins = [1], amount = 0
输出：0

示例 4：

输入：coins = [1], amount = 1
输出：1

示例 5：

输入：coins = [1], amount = 2
输出：2

 

提示：

• 1 <= coins.length <= 12
• 1 <= coins[i] <= 231 - 1
• 0 <= amount <= 104

解法

动态规划。

类似完全背包的思路，硬币数量不限，求凑成总金额所需的最少的硬币个数。

Python3

class Solution:
    def coinChange(self, coins: List[int], amount: int) -> int:
        dp = [amount + 1] * (amount + 1)
        dp[0] = 0
        for coin in coins:
            for j in range(coin, amount + 1):
                dp[j] = min(dp[j], dp[j - coin] + 1)
        return -1 if dp[-1] > amount else dp[-1]

Java

class Solution {
    public int coinChange(int[] coins, int amount) {
        int m = coins.length;
        int[][] dp = new int[m + 1][amount + 1];
        for (int i = 0; i <= m; ++i) {
            Arrays.fill(dp[i], amount + 1);
        }
        dp[0][0] = 0;
        for (int i = 1; i <= m; ++i) {
            int v = coins[i - 1];
            for (int j = 0; j <= amount; ++j) {
                for (int k = 0; k * v <= j; ++k) {
                    dp[i][j] = Math.min(dp[i][j], dp[i - 1][j - k * v] + k);
                }
            }
        }
        return dp[m][amount] > amount ? - 1 : dp[m][amount];
    }
}

下面对 k 这层循环进行优化：

由于：

• dp[i][j] = min(dp[i - 1][j], dp[i - 1][j - v] + 1, dp[i - 1][j - 2v] + 2, ... , dp[i - 1][j - kv] + k)
• dp[i][j - v] = min( dp[i - 1][j - v], dp[i - 1][j - 2v] + 1, ... , dp[i - 1][j - kv] + k - 1)

因此 dp[i][j] = min(dp[i - 1][j], dp[i][j - v] + 1)。

class Solution {
    public int coinChange(int[] coins, int amount) {
        int m = coins.length;
        int[][] dp = new int[m + 1][amount + 1];
        for (int i = 0; i <= m; ++i) {
            Arrays.fill(dp[i], amount + 1);
        }
        dp[0][0] = 0;
        for (int i = 1; i <= m; ++i) {
            int v = coins[i - 1];
            for (int j = 0; j <= amount; ++j) {
                dp[i][j] = dp[i - 1][j];
                if (j >= v) {
                    dp[i][j] = Math.min(dp[i][j], dp[i][j - v] + 1);
                }
            }
        }
        return dp[m][amount] > amount ? - 1 : dp[m][amount];
    }
}

空间优化：

class Solution {
    public int coinChange(int[] coins, int amount) {
        int[] dp = new int[amount + 1];
        Arrays.fill(dp, amount + 1);
        dp[0] = 0;
        for (int coin : coins) {
            for (int j = coin; j <= amount; j++) {
                dp[j] = Math.min(dp[j], dp[j - coin] + 1);
            }
        }
        return dp[amount] > amount ? -1 : dp[amount];
    }
}

JavaScript

/**
 * @param {number[]} coins
 * @param {number} amount
 * @return {number}
 */
var coinChange = function (coins, amount) {
    let dp = Array(amount + 1).fill(amount + 1);
    dp[0] = 0;
    for (const coin of coins) {
        for (let j = coin; j <= amount; ++j) {
            dp[j] = Math.min(dp[j], dp[j - coin] + 1);
        }
    }
    return dp[amount] > amount ? -1 : dp[amount];
};

C++

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

Go

func coinChange(coins []int, amount int) int {
	dp := make([]int, amount+1)
	for i := 1; i <= amount; i++ {
		dp[i] = amount + 1
	}
	for _, coin := range coins {
		for j := coin; j <= amount; j++ {
			dp[j] = min(dp[j], dp[j-coin]+1)
		}
	}
	if dp[amount] > amount {
		return -1
	}
	return dp[amount]
}

func min(a, b int) int {
	if a < b {
		return a
	}
	return b
}