README.md

322. 零钱兑换

English Version

题目描述

给你一个整数数组 coins ，表示不同面额的硬币；以及一个整数 amount ，表示总金额。

计算并返回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

 

提示：

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

解法

方法一：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。

定义 $dp[i][j]$ 表示从前 $i$ 种硬币选出总金额为 $j$ 所需的最少硬币数。

那么有：

$$ dp[i][j] = \min(dp[i - 1][j], dp[i - 1][j - v] + 1, dp[i - 1][j - 2\times v] + 2, ... , dp[i - 1][j - k\times v] + k) $$

令 $j=j-v$，则有：

$$ dp[i][j - v] = \min( dp[i - 1][j - v], dp[i - 1][j - 2\times v] + 1, ... , dp[i - 1][j - k\times v] + k - 1) $$

因此，我们可以得到状态转移方程：

$$ dp[i][j] = \min(dp[i - 1][j], dp[i][j - v] + 1) $$

时间复杂度 $O(m\times n)$，空间复杂度 $O(m\times n)$。其中 $m$ 和 $n$ 分别为硬币数量和总金额。

Python3

动态规划——完全背包问题朴素做法：

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

动态规划——完全背包问题空间优化：

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) {
                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];
    }
}

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
}

TypeScript

function coinChange(coins: number[], amount: number): number {
    let dp = new 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];
}

Rust

impl Solution {
    pub fn coin_change(coins: Vec<i32>, amount: i32) -> i32 {
        let n = coins.len();
        let amount = amount as usize;
        let mut dp = vec![amount + 1; amount + 1];
        dp[0] = 0;
        for i in 1..=amount {
            for j in 0..n {
                let coin = coins[j] as usize;
                if coin <= i {
                    dp[i] = dp[i].min(dp[i - coin] + 1);
                }
            }
        }
        if dp[amount] > amount {
            -1
        } else {
            dp[amount] as i32
        }
    }
}

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