[LeetCode]322. Coin Change

Description:

You are given coins of different denominations and a total amount of money amount. Write a function to compute the fewest number of coins that you need to make up that amount. If that amount of money cannot be made up by any combination of the coins, return -1.

Example 1:
coins = [1, 2, 5], amount = 11
return 3 (11 = 5 + 5 + 1)

Example 2:
coins = [2], amount = 3
return -1.

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

————————————————————————————————————————————————————————————————

Solution:

題意：湊金錢面額。

思路：動態規劃。這裏的動態規劃數組不是基於面值數量的數組，而是基於需要湊整的面額數組。

class Solution {
public:
    int coinChange(vector<int>& coins, int amount) {
        vector<int> dp(amount + 1, amount + 1);
        dp[0] = 0;
        for (int i = 1; i <= amount; i++) {
            for (vector<int>::iterator j = coins.begin(); j != coins.end(); j++) {
                if (*j <= i) {
                    // 如果無法組成amount，dp[amount] = min(amount + 1, amount + 1 + 1) = amount + 1
                    dp[i] = min(dp[i], dp[i - *j] + 1);
                }
            }
        }
        return dp[amount] > amount ? -1 : dp[amount];
    }
};