README_EN.md

322. Coin Change

中文文档

Description

You are given an integer array coins representing coins of different denominations and an integer amount representing a total amount of money.

Return 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.

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

 

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

Example 3:

Input: coins = [1], amount = 0
Output: 0

Example 4:

Input: coins = [1], amount = 1
Output: 1

Example 5:

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

 

Constraints:

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

Solutions

Similar to the idea of ​​a complete backpack, there is no limit to the number of coins. Find the minimum number of coins required to make up the total amount.

Python3

class Solution:
    def coinChange(self, coins: List[int], amount: int) -> int:
        dp = [amount + 1 for i in range(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) {
    var dp = Array(amount + 1).fill(amount + 1);
    dp[0] = 0;
    for (var i = 1; i <= amount; i++) {
        for (var j = 0; j < coins.length; j++) {
            if (coins[j] <= i) {
                dp[i] = Math.min(dp[i], dp[i - coins[j]] + 1);
            }
        }
    }

    return dp[amount] > amount ? -1 : dp[amount];
};

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