class Solution {
public:
int change(int amount, vector<int>& coins) {
}
};
518. 零钱兑换 II
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 <= 3001 <= coins[i] <= 5000coins 中的所有值 互不相同0 <= amount <= 5000原站题解
rust 解法, 执行用时: 2 ms, 内存消耗: 2 MB, 提交时间: 2024-03-25 09:28:30
impl Solution {
pub fn change(amount: i32, coins: Vec<i32>) -> i32 {
let mut dp = vec![0; amount as usize + 1];
dp[0] = 1;
for coin in coins {
for i in coin..=amount {
dp[i as usize] += dp[(i - coin) as usize];
}
}
dp[amount as usize]
}
}
php 解法, 执行用时: 18 ms, 内存消耗: 20.1 MB, 提交时间: 2024-03-25 09:27:47
class Solution {
/**
* @param Integer $amount
* @param Integer[] $coins
* @return Integer
*/
function change($amount, $coins) {
$dp = array_fill(0, $amount + 1, 0);
$dp[0] = 1;
foreach ($coins as $coin) {
for ($i = $coin; $i <= $amount; $i++) {
$dp[$i] += $dp[$i - $coin];
}
}
return $dp[$amount];
}
}
javascript 解法, 执行用时: 60 ms, 内存消耗: 49.7 MB, 提交时间: 2024-03-25 09:27:08
/**
* @param {number} amount
* @param {number[]} coins
* @return {number}
*/
var change = function(amount, coins) {
const 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[amount];
};
cpp 解法, 执行用时: 6 ms, 内存消耗: 8.3 MB, 提交时间: 2024-03-25 09:26:13
class Solution {
public:
int change(int amount, vector<int>& coins) {
std::vector<int> dp(amount + 1, 0);
dp[0] = 1;
for (int coin : coins) {
for (int i = coin; i <= amount; ++i) {
dp[i] += dp[i - coin];
}
}
return dp[amount];
}
};
java 解法, 执行用时: 2 ms, 内存消耗: 39.9 MB, 提交时间: 2024-03-25 09:25:36
class Solution {
public int change(int amount, int[] coins) {
int[] dp = new int[amount + 1];
dp[0] = 1;
for (int coin : coins) {
for (int i = coin; i <= amount; i++) {
dp[i] += dp[i - coin];
}
}
return dp[amount];
}
}
golang 解法, 执行用时: 4 ms, 内存消耗: 2.2 MB, 提交时间: 2022-11-09 15:00:04
func change(amount int, coins []int) int {
dp := make([]int, amount+1)
dp[0] = 1
for _, coin := range coins {
for i := coin; i <= amount; i++ {
dp[i] += dp[i-coin]
}
}
return dp[amount]
}
python3 解法, 执行用时: 232 ms, 内存消耗: 15.2 MB, 提交时间: 2022-11-09 14:59:38
class Solution:
def change(self, amount: int, coins: List[int]) -> int:
'''
dp[x]: 总额为x的货币组合种类
dp[0] = 1
dp[x] =
'''
dp = [0] * (amount + 1)
dp[0] = 1
for coin in coins:
for i in range(coin, amount+1):
dp[i] += dp[i-coin]
return dp[amount]