Question

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(完全背包)

我们定义 $f[i][j]$ 表示使用前 $i$ 种硬币，凑出金额 $j$ 的最少硬币数。初始时 $f[0][0] = 0$，其余位置的值均为正无穷。

我们可以枚举使用的最后一枚硬币的数量 $k$，那么有：

$$ f[i][j] = \min(f[i - 1][j], f[i - 1][j - x] + 1, \cdots, f[i - 1][j - k \times x] + k) $$

其中 $x$ 表示第 $i$ 种硬币的面值。

不妨令 $j = j - x$，那么有：

$$ f[i][j - x] = \min(f[i - 1][j - x], f[i - 1][j - 2 \times x] + 1, \cdots, f[i - 1][j - k \times x] + k - 1) $$

将二式代入一式，我们可以得到以下状态转移方程：

$$ f[i][j] = \min(f[i - 1][j], f[i][j - x] + 1) $$

最后答案即为 $f[m][n]$。

时间复杂度 $O(m \times n)$，空间复杂度 $O(m \times n)$。其中 $m$ 和 $n$ 分别为硬币的种类数和总金额。

注意到 $f[i][j]$ 只与 $f[i - 1][j]$ 和 $f[i][j - x]$ 有关，因此我们可以将二维数组优化为一维数组，空间复杂度降为 $O(n)$。

相似题目：

• 279. 完全平方数

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

class Solution {
  public int coinChange(int[] coins, int amount) {
    final int inf = 1 << 30;
    int m = coins.length;
    int n = amount;
    int[][] f = new int[m + 1][n + 1];
    for (var g : f) {
      Arrays.fill(g, inf);
    }
    f[0][0] = 0;
    for (int i = 1; i <= m; ++i) {
      for (int j = 0; j <= n; ++j) {
        f[i][j] = f[i - 1][j];
        if (j >= coins[i - 1]) {
          f[i][j] = Math.min(f[i][j], f[i][j - coins[i - 1]] + 1);
        }
      }
    }
    return f[m][n] >= inf ? -1 : f[m][n];
  }
}

class Solution {
public:
  int coinChange(vector<int>& coins, int amount) {
    int m = coins.size(), n = amount;
    int f[m + 1][n + 1];
    memset(f, 0x3f, sizeof(f));
    f[0][0] = 0;
    for (int i = 1; i <= m; ++i) {
      for (int j = 0; j <= n; ++j) {
        f[i][j] = f[i - 1][j];
        if (j >= coins[i - 1]) {
          f[i][j] = min(f[i][j], f[i][j - coins[i - 1]] + 1);
        }
      }
    }
    return f[m][n] > n ? -1 : f[m][n];
  }
};

func coinChange(coins []int, amount int) int {
  m, n := len(coins), amount
  f := make([][]int, m+1)
  const inf = 1 << 30
  for i := range f {
    f[i] = make([]int, n+1)
    for j := range f[i] {
      f[i][j] = inf
    }
  }
  f[0][0] = 0
  for i := 1; i <= m; i++ {
    for j := 0; j <= n; j++ {
      f[i][j] = f[i-1][j]
      if j >= coins[i-1] {
        f[i][j] = min(f[i][j], f[i][j-coins[i-1]]+1)
      }
    }
  }
  if f[m][n] > n {
    return -1
  }
  return f[m][n]
}

function coinChange(coins: number[], amount: number): number {
  const m = coins.length;
  const n = amount;
  const f: number[][] = Array(m + 1)
    .fill(0)
    .map(() => Array(n + 1).fill(1 << 30));
  f[0][0] = 0;
  for (let i = 1; i <= m; ++i) {
    for (let j = 0; j <= n; ++j) {
      f[i][j] = f[i - 1][j];
      if (j >= coins[i - 1]) {
        f[i][j] = Math.min(f[i][j], f[i][j - coins[i - 1]] + 1);
      }
    }
  }
  return f[m][n] > n ? -1 : f[m][n];
}

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

/**
 * @param {number[]} coins
 * @param {number} amount
 * @return {number}
 */
var coinChange = function (coins, amount) {
  const m = coins.length;
  const n = amount;
  const f = Array(m + 1)
    .fill(0)
    .map(() => Array(n + 1).fill(1 << 30));
  f[0][0] = 0;
  for (let i = 1; i <= m; ++i) {
    for (let j = 0; j <= n; ++j) {
      f[i][j] = f[i - 1][j];
      if (j >= coins[i - 1]) {
        f[i][j] = Math.min(f[i][j], f[i][j - coins[i - 1]] + 1);
      }
    }
  }
  return f[m][n] > n ? -1 : f[m][n];
};

方法二

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

class Solution {
  public int coinChange(int[] coins, int amount) {
    final int inf = 1 << 30;
    int n = amount;
    int[] f = new int[n + 1];
    Arrays.fill(f, inf);
    f[0] = 0;
    for (int x : coins) {
      for (int j = x; j <= n; ++j) {
        f[j] = Math.min(f[j], f[j - x] + 1);
      }
    }
    return f[n] >= inf ? -1 : f[n];
  }
}

class Solution {
public:
  int coinChange(vector<int>& coins, int amount) {
    int n = amount;
    int f[n + 1];
    memset(f, 0x3f, sizeof(f));
    f[0] = 0;
    for (int x : coins) {
      for (int j = x; j <= n; ++j) {
        f[j] = min(f[j], f[j - x] + 1);
      }
    }
    return f[n] > n ? -1 : f[n];
  }
};

func coinChange(coins []int, amount int) int {
  n := amount
  f := make([]int, n+1)
  for i := range f {
    f[i] = 1 << 30
  }
  f[0] = 0
  for _, x := range coins {
    for j := x; j <= n; j++ {
      f[j] = min(f[j], f[j-x]+1)
    }
  }
  if f[n] > n {
    return -1
  }
  return f[n]
}

function coinChange(coins: number[], amount: number): number {
  const n = amount;
  const f: number[] = Array(n + 1).fill(1 << 30);
  f[0] = 0;
  for (const x of coins) {
    for (let j = x; j <= n; ++j) {
      f[j] = Math.min(f[j], f[j - x] + 1);
    }
  }
  return f[n] > n ? -1 : f[n];
}

/**
 * @param {number[]} coins
 * @param {number} amount
 * @return {number}
 */
var coinChange = function (coins, amount) {
  const n = amount;
  const f = Array(n + 1).fill(1 << 30);
  f[0] = 0;
  for (const x of coins) {
    for (let j = x; j <= n; ++j) {
      f[j] = Math.min(f[j], f[j - x] + 1);
    }
  }
  return f[n] > n ? -1 : f[n];
};