Question

518. 零钱兑换 II

English Version

题目描述

给你一个整数数组 coins 表示不同面额的硬币，另给一个整数 amount 表示总金额。

请你计算并返回可以凑成总金额的硬币组合数。如果任何硬币组合都无法凑出总金额，返回 0 。

假设每一种面额的硬币有无限个。 

题目数据保证结果符合 32 位带符号整数。

 

示例 1：

输入：amount = 5, coins = [1, 2, 5]
输出：4
解释：有四种方式可以凑成总金额：
5=5
5=2+2+1
5=2+1+1+1
5=1+1+1+1+1

示例 2：

输入：amount = 3, coins = [2]
输出：0
解释：只用面额 2 的硬币不能凑成总金额 3 。

示例 3：

输入：amount = 10, coins = [10] 
输出：1

 

提示：

• 1 <= coins.length <= 300
• 1 <= coins[i] <= 5000
• coins 中的所有值 互不相同
• 0 <= amount <= 5000

解法

方法一

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

class Solution {
  public int change(int amount, int[] coins) {
    int m = coins.length;
    int[][] dp = new int[m + 1][amount + 1];
    dp[0][0] = 1;
    for (int i = 1; i <= m; ++i) {
      for (int j = 0; j <= amount; ++j) {
        for (int k = 0; k * coins[i - 1] <= j; ++k) {
          dp[i][j] += dp[i - 1][j - coins[i - 1] * k];
        }
      }
    }
    return dp[m][amount];
  }
}

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

func change(amount int, coins []int) int {
  dp := make([]int, amount+1)
  dp[0] = 1
  for _, coin := range coins {
    for j := coin; j <= amount; j++ {
      dp[j] += dp[j-coin]
    }
  }
  return dp[amount]
}

function change(amount: number, coins: number[]): number {
  let dp = new Array(amount + 1).fill(0);
  dp[0] = 1;
  for (let coin of coins) {
    for (let i = coin; i <= amount; ++i) {
      dp[i] += dp[i - coin];
    }
  }
  return dp.pop();
}

方法二

class Solution {
  public int change(int amount, int[] coins) {
    int m = coins.length;
    int[][] dp = new int[m + 1][amount + 1];
    dp[0][0] = 1;
    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] += dp[i][j - v];
        }
      }
    }
    return dp[m][amount];
  }
}

方法三

class Solution {
  public int change(int amount, int[] coins) {
    int[] dp = new int[amount + 1];
    dp[0] = 1;
    for (int coin : coins) {
      // 顺序遍历，0-1背包问题是倒序遍历
      for (int j = coin; j <= amount; j++) {
        dp[j] += dp[j - coin];
      }
    }
    return dp[amount];
  }
}