leetcode刷题. 322. 动态规划的解法复习

322. 零钱兑换

注：此题还是比较经典的！

动态规划 + 贪心

int coinChange(vector<int>& coins, int amount) {

	// 这里注意开辟空间的大小，防止溢出
	int F_[10000] = { 0 };

	for (int s = 1; s <= amount; s++) {

		int idx_ = 0;
		int sMin_ = INT_MAX;

		for (int i = 0; i < coins.size(); i++) {

			if (s - coins[i] < 0)
				continue;

			idx_ = s - coins[i];
			if (F_[idx_] == -1)
				continue;

			if (F_[idx_] + 1 < sMin_)
				sMin_ = F_[idx_] + 1;
		}

		if (sMin_ == INT_MAX) {

			F_[s] = -1;
			continue;
		}
		F_[s] = sMin_;
	}
	return F_[amount];
}