[LeetCode] 518. Coin Change 2

You are given coins of different denominations and a total amount of money. Write a function to compute the number of combinations that make up that amount. You may assume that you have infinite number of each kind of coin.

Example 1:

Input: amount = 5, coins = [1, 2, 5]
Output: 4
Explanation: there are four ways to make up the amount:
5=5
5=2+2+1
5=2+1+1+1
5=1+1+1+1+1

Example 2:

Input: amount = 3, coins = [2]
Output: 0
Explanation: the amount of 3 cannot be made up just with coins of 2.

Example 3:

Input: amount = 10, coins = [10] 
Output: 1 

Note:

You can assume that

• 0 <= amount <= 5000
• 1 <= coin <= 5000
• the number of coins is less than 500
• the answer is guaranteed to fit into signed 32-bit integer

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。 

思路依然是动态规划，同时这是一道完全背包问题。背包问题的定义是一个组合的不同排列在结果集里只出现过一次，比如 [1, 1, 2] 和 [1, 2, 1]，只能算是一种情况。

dp[i] 的定义是当 amount = i 的时候，到底有几种硬币组合。以基本情况没有硬币开始组合数量。初始化 dp[0] = 1，其余等于 0。遍历所有硬币面值

• 对于每个硬币coin，我们将从金额 coin 遍历到 amount
• 对于每个 x，计算组合数：dp[x] += dp[x - coin]
• 返回 dp[amount]

时间O(n^2)

空间O(n)

Java实现

class Solution {
    public int change(int amount, int[] coins) {
        int[] dp = new int[amount + 1];
        dp[0] = 1;
        for (int coin : coins) {
            for (int x = coin; x < amount + 1; x++) {
                dp[x] += dp[x - coin];
            }
        }
        return dp[amount];
    }
}

 

JavaScript实现

/**
 * @param {number} amount
 * @param {number[]} coins
 * @return {number}
 */
var change = function(amount, coins) {
    let dp = new Array(amount + 1).fill(0);
    dp[0] = 1;
    for (let coin of coins) {
        for (let x = coin; x < amount + 1; x++) {
            dp[x] += dp[x - coin];
        }
    }
    return dp[amount];
};

 

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