刷题322. Coin Change

一、题目说明

题目322. Coin Change，给定一组不同面值的硬币，计算给定的总金额可以用硬币凑成的最小数量。难度是Medium！

二、我的解答

这个题目，思考了一下，和前面的279. Perfect Squares有点类似，属于求最优解的问题。解答方法无外乎用递归，或者dp。但是这个没做出来，由于没有找到最优子结构。网上找到的代码：

class Solution{
	public:
		//dfs + memorization
		int coinChange(vector<int>& coins,int amount){
			vector<int> memo(amount+1,-2);
			return dfs(coins,amount,memo);
		}
		
		int dfs(vector<int>& coins,int amount,vector<int>& memo){
			if(amount == 0) return 0;
			if(memo[amount] != -2){
				return memo[amount];
			}
			int ans = INT_MAX;
			for(int coin: coins){
				if(amount-coin<0) continue;
				int subProc = dfs(coins,amount-coin,memo);
				if(subProc == -1){
					continue;
				}
				ans = min(ans,subProc+1);
			}
			memo[amount] = (ans == INT_MAX) ? -1 : ans;
			return memo[amount];
		}
};

性能：

Runtime: 80 ms, faster than 22.84% of C++ online submissions for Coin Change.
Memory Usage: 14.5 MB, less than 27.45% of C++ online submissions for Coin Change.

三、优化措施

用dp写：

class Solution{
	public:
		//dp solution: dp[i] means the minimum num of coins used
		int coinChange(vector<int>& coins,int amount){
			vector<int> dp(amount+1,amount+1);
			dp[0] = 0;
			int len = coins.size();
			for(int i=1;i<=amount;i++){
				for(int j=0;j<len;j++){
					if(coins[j]<=i){
						dp[i] = min(dp[i],dp[i-coins[j]] + 1);
					}
				}
			}
			if(dp[amount]==amount+1){
				return -1;
			}else return dp[amount];
		}
};

性能如下：

Runtime: 48 ms, faster than 70.77% of C++ online submissions for Coin Change.
Memory Usage: 12.6 MB, less than 86.27% of C++ online submissions for Coin Change.