README_CN.md

518. 零钱兑换 II

给定不同面额的硬币和一个总金额。写出函数来计算可以凑成总金额的硬币组合数。假设每一种面额的硬币有无限个。

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

注意:

你可以假设：

• 0 <= amount (总金额) <= 5000
• 1 <= coin (硬币面额) <= 5000
• 硬币种类不超过 500 种
• 结果符合 32 位符号整数

题解 (Rust)

1. 动态规划

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 as usize)..=(amount as usize) {
                dp[i] += dp[i - coin as usize];
            }
        }

        dp[amount as usize]
    }
}