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

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[amount] > amount else dp[amount]

Java

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];
    }
};