518.Coin Change II

tags: Medium,Array,DP

518. Coin Change II

題目描述

You are given an integer array coins representing coins of different denominations and an integer amount representing a total amount of money.

Return the number of combinations that make up that amount. If that amount of money cannot be made up by any combination of the coins, return 0.

You may assume that you have an infinite number of each kind of coin.

The answer is guaranteed to fit into a signed 32-bit integer.

範例

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

Constraints:

• 1 <= coins.length <= 300
• 1 <= coins[i] <= 5000
• All the values of coins are unique.
• 0 <= amount <= 5000

解答

TypeScript

function change(amount: number, coins: number[]): number {
  const dp: number[] = new Array(amount + 1).fill(0);
  dp[0] = 1;

  for (const coin of coins) {
    for (let i = coin; i <= amount; i++) {
      dp[i] += dp[i - coin];
    }
  }

  return dp[amount];
}

思路：DP

1. 建立一個 DP table, dp[i] 代表組成 i 元的組合數
2. dp[0] = 1, 代表組成 0 元的組合數為 1 (都不選)
3. dp[i - coin] 代表組成 i - coin 元的組合數，因為當我們在湊出 i - coin 的組合數基礎上，再加上 coin 就可以湊出 i 元的一種新組合
   以 coins = [1, 2], amount = 3 為例：
   dp[0] = 1 => []
   dp[1] = dp[1 - 1] = 1 => [1]
   dp[2] = dp[2 - 1] + dp[2 - 2] = 2 => [1, 1], [2]
   dp[3] = dp[3 - 1] + dp[3 - 2] = 3 => [1, 1, 1], [1, 2], [2, 1]
4. 遍歷 coins，並從 coin 開始遍歷 dp，直到 amount，然後更新 dp[i] 把每種 dp[i - coin] 的組合數加上去
5. 回傳 dp[amount]
   SheepFri, Aug 11, 2023

C++

class Solution {
public:
    int change(int amount, vector<int>& coins) {
        ios_base::sync_with_stdio(0); cin.tie(0);

        int numCoins = coins.size();
        vector<int> dp(amount + 1, 0);
        dp[0] = 1;
        for (const int coin : coins) {
            for (int currHold = coin; currHold <= amount; currHold ++) {
                dp[currHold] += dp[currHold - coin];
            }
        }
        return dp[amount];
    }
};

Jerry Wu11 August, 2023

Reference

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