动态规划，3个硬币的问题：https://leetcode.com/problems/coin-change/
解答思路：https://leetcode.com/problems/coin-change/discuss/77360/C%2B%2B-O(n*amount)-time-O(amount)-space-DP-solution
代码：
class Solution {
public:
int coinChange(vector<int>& coins, int amount) {
int Max = amount + 1;
vector<int> dp(amount + 1, Max);
dp[0] = 0;
for (int i = 1; i <= amount; i++) {
for (int j = 0; j < coins.size(); j++) {
if (coins[j] <= i) {
dp[i] = min(dp[i], dp[i - coins[j]] + 1);
}
}
}
return dp[amount] > amount ? -1 : dp[amount]; // 如果不能凑够目标金额，需要返回-1
}
};
