Given coins of different denominations and a total amount. Write a function to calculate the minimum number of coins required to make up the total amount. If no coin combination can make up the total amount, return -1.
You can think that the number of each coin is unlimited.
cpp implementation
class Solution {
public:
int coinChange(vector<int>& coins, int amount) {
// 方法一：动态规划
// 等于钞票面值的最优解
// 找出最小的i-coins[j]的金额就是最优解dp[i-coins[j]]+1
// 1 2 5 7 10 //14
// 0 1 2 3 4 5 6 7 8 9 10 11 12 13
// dp 0 1 1 2 2 1 2 1 2 2 1 2 3 2
// 初始化数组dp, 大小是amount+1, 全部元素初始化为-1
vector<int> dp(amount+1, -1);
// 金额0的最优解dp[0]=0
dp[0] = 0;
// 变量i从1循环到amount，一次计算金额1到amount的最优解
for(int i=1; i<=amount; i++){
// 对于每一个金额i, 使用变量j遍历面值c
for(int j=0; j<coins.size(); j++){
// 所有与小于等于i的面值coins[j], 如果金额i-coins[j]有最优解
if(coins[j]<=i && dp[i-coins[j]]!=-1){
// 如果当前金额还未计算或者dp[i]比正在计算的最优解大(避免重复计算)
if(dp[i]==-1||dp[i]>dp[i-coins[j]]+1){
// 更新dp[i]
dp[i] = dp[i-coins[j]] + 1;
}
}
}
}
return dp[amount];
}
};。
py implementation
class Solution:
def coinChange(self, coins: List[int], amount: int) -> int:
# 初始化dp数组为-1
dp = []
for i in range(amount+1):
dp.append(-1)
dp[0] = 0
# 遍历变量i从1到amount,依次计算金额1到amount的最优解
for i in range(1, amount+1):
for j in range(len(coins)):
if coins[j] <= i and dp[i-coins[j]] != -1:
if dp[i] == -1 or dp[i] > dp[i-coins[j]]+1:
dp[i] = dp[i-coins[j]] + 1
return dp[amount]
Source of test questions: LeetCode
Link: https://leetcode-cn.com/problems/coin-change