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
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
来源:力扣(LeetCode)
链接:https://leetcode-cn.com/problems/coin-change
参考:
- https://leetcode-cn.com/problems/coin-change/solution/dai-ma-sui-xiang-lu-dai-ni-xue-tou-wan-q-80r7/
python
# 0322.零钱兑换
class Solution:
def coinChange(self, coins: [int], amount: int) -> int:
"""
动态规划,完全背包,先遍历物品再遍历背包
:param coins:
:param amount:
:return:
"""
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)
print(" dp[%d]: %d |" % (j, dp[j]), end=' ')
return dp[amount] if dp[amount] < amount+1 else -1
def coinChange2(self, coins: [int], amount: int) -> int:
"""
动态规划,完全背包,先背包再物品
:param coins:
:param amount:
:return:
"""
dp = [amount+1] * (amount+1)
dp[0] = 0
for j in range(1, amount+1):
for i in range(len(coins)):
if j >= coins[i]:
dp[j] = min(dp[j], dp[j-coins[i]]+1)
return dp[amount] if dp[amount] < amount+1 else -1
if __name__ == "__main__":
coins = [346,29,395,188,155,109]
test = Solution()
test.coinChange(coins, 9401)
golang
package dynamicPrograming
import "math"
// 动态规划-完全背包-先物品再背包
func coinChange(coins []int, amount int) int {
dp := make([]int, amount+1)
dp[0] = 0
for j:=1;j<=amount;j++ {
dp[j] = math.MaxInt32
}
// 遍历物品
for i:=0;i<len(coins);i++ {
for j:=coins[i];j<=amount;j++ {
if dp[j-coins[i]] != math.MaxInt32 {
dp[j] = min(dp[j], dp[j-coins[i]]+1)
}
}
}
if dp[amount] == math.MaxInt32 {
return -1
}
return dp[amount]
}
// 动态规划-完全背包-先背包再物品
func coinChange2(coins []int, amount int) int {
dp := make([]int, amount+1)
dp[0] = 0
for j:=1;j<=amount;j++ {
dp[j] = math.MaxInt32
for i:=0;i<len(coins);i++ {
if j >= coins[i] && dp[j-coins[i]] != math.MaxInt32 {
dp[j] = min(dp[j], dp[j-coins[i]]+1)
}
}
}
if dp[amount] == math.MaxInt32 {
return -1
}
return dp[amount]
}
func min(a,b int) int {
if a < b {
return a
}
return b
}