LeetCode: Coin Change

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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.

题目描述：给定无限个数量的不同面值的硬币和一个总额，返回满足总额的最少硬币数量

解决思路1：动态规划

f(n) = min(INF, f(n - i) + 1) // 要么不能满足，要么是选用当前可以用的一个硬币

代码如下：

int coinChange(int* coins, int coinsSize, int amount) {        int *num = (int*)malloc((amount + 1)* sizeof(int));        for (int i = 0; i < amount; ++i) {        num[i] = -1;    }    num[0] = 0;        for (int i = 1; i <= amount; ++i) {                int min_amount = 2 * amount;         for (int j = 0; j < coinsSize; ++j) {            if (i - coins[j] >= 0 && num[i - coins[j]] >= 0) {                if (min_amount >= 1 + num[i - coins[j]]) {                    min_amount = 1 + num[i - coins[j]];                }            }        }        if (min_amount == 2 * amount) {            num[i] = -1;        }        else {            num[i] = min_amount;        }    }        return num[amount];    }

用这种方法后，发现时间代价很高，说明可以优化

解决思路2：

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