【Day 64 】2024-06-10 - 518. 零钱兑换 II

[azl397985856]

518. 零钱兑换 II

入选理由

暂无

题目地址

https://leetcode-cn.com/problems/coin-change-2/

前置知识

暂无

题目描述

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

[Martina001]

经典完全背包问题 class Solution { public int change(int amount, int[] coins) { int[] f = new int[amount + 1]; f[0] = 1; for (int x : coins) { for (int c = x; c <= amount; c++) { f[c] += f[c - x]; } } return f[amount]; } }

[franklinsworld666]

代码

class Solution:
    def change(self, amount: int, coins: List[int]) -> int:
        dp = [0] * (amount + 1)
        dp[0] = 1
        for coin in coins:
            for i in range(1, amount + 1):
                if coin <= i:
                    dp[i] += dp[i - coin]
        return dp[amount]

[xy147]

js代码

var change = function (amount, coins) {
  const dp = Array.from({ length: 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];
};

复杂度分析

时间复杂度：O(m*n) 空间复杂度：O(m)

[hillsonziqiu]

思路

代码

/**
 * @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 (const item of coins) {
        for (let i = item; i <= amount; i++) {
            dp[i] += dp[i - item];
        }
    }
    return dp[amount]
};

复杂度分析

• 时间复杂度：O(amount * n);
• 空间复杂度：O(amount);

[Dtjk]

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]; } }

[Hermione666]

class Solution: def change(self, amount: int, coins: list[int]) -> int: 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]