Coin Change

题目描述：

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.

算法：

明显地，本题我们应该使用动态规划来解决。我们用一个数组dp来保存状态，dp[i]的值的含义是，当硬币的面额总和为i时，所需要的最少的硬币数。这样状态转移方程为dp[j]=min(dp[j],dp[j-coins[i]]+1)，该方程的含义是：当我们要求解面额总和j所需的最少硬币数时，我们需要从下面两个值中取一个最小值，一个是当前构成面额为j所需的硬币数，另一个是构成面额为j-coins[i]所需的硬币数再加上面额为coins[i]的那个硬币。则最终的答案是dp[amount]。这里，我们首先将数组dp全部初始化为INT_MAX，如果最终dp[amount]的值为INT_MAX，那么表示无法构成该面额。代码实现如下：

int coinChange(vector<int>& coins, int amount) {vector<int> dp(amount + 1, INT_MAX-100);dp[0] = 0;for (int i = 0; i < coins.size(); i++){if (coins[i]<=amount)dp[coins[i]] = 1;}for (int i = 0; i < coins.size(); i++){for (int j = 0; j <= amount; j++){if (j >= coins[i]){dp[j] = min(dp[j], dp[j - coins[i]] + 1);//cout << j << ' ' << dp[j] << endl;}}}if (dp[amount] == INT_MAX - 100)return -1;return dp[amount];}