322. Coin Change

https://leetcode.com/problems/coin-change/

solution

为什么不能用贪心，尽量选面值最高的。假设有以下硬币面值：[1, 3, 4]，需要找零6元

class Solution:
    def coinChange(self, coins: List[int], amount: int) -> int:
        dp = [float('inf')] * (amount + 1)
        dp[0] = 0

        for i in range(1, amount+1):
            for coin in coins:
                if i - coin >= 0:
                    dp[i] = min(dp[i], dp[i-coin]+1)
        return dp[-1] if dp[-1] != float('inf') else -1

• 完全背包

  • dp[j]：凑足总额为j所需钱币的最少个数为dp[j]. 先物品在背包理解成,

class Solution:
    def coinChange(self, coins: List[int], amount: int) -> int:
        dp = [float('inf') for _ in range(amount + 1)]  # 恰好装满类的背包问题初始化
        dp[0] = 0

        for i in coins:  # 先物品
            for j in range(i, amount+1):  # 再背包(或者不能说背包，而是背包的约束), 背包大于物品
                if dp[j-i] != float('inf'):
                    dp[j] = min(dp[j], dp[j-i] + 1)

        if dp[amount] == float('inf'):
            return -1
        return dp[amount]

时间复杂度：O(∣coins∣∣amount∣) 空间复杂度：O(∣amount∣)

• Recursion

# https://www.geeksforgeeks.org/coin-change-dp-7/
# https://leetcode.com/problems/coin-change/solutions/77409/evolve-from-brute-force-to-optimal-a-review-of-all-solutions/

时间复杂度：O() 空间复杂度：O()

follow up

518. Coin Change II