Leetcode 322 Coin Change DP(Leetcode 322 Coin Change DP)

You are given an integer array representing coins of different denominations and an integer representing a total amount of money.

coins
amount

Return 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

You may assume that you have an infinite number of each kind of coin.

Solution

设 \(dp[i]\) 表示 \(amount = i\) 的最小组成数,转移的话很显然就是枚举每一个小于 \(amount\) 的钱币:

class Solution {
private:
    int dp[10004];
    
public:
    int coinChange(vector<int>& coins, int amount) {
        if(amount==0)return 0;
        int n = coins.size();
        int MAX = 9999999;
        for(int i=0;i<=amount;i++)dp[i] = MAX;
        for(int j=0;j<n;j++){
            if(amount>=coins[j])dp[coins[j]]=1;
        }
        
        for(int i=1;i<=amount;i++){
            for(int j=0;j<n;j++){
                if(coins[j]<=i){
                    dp[i] = min(dp[i], 1 + dp[i-coins[j]] );
                }
            }
        }
        if(dp[amount]>amount)return -1;
        else return dp[amount];
    }
};
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You are given an integer array representing coins of different denominations and an integer representing a total amount of money.

coins
amount

Return 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

You may assume that you have an infinite number of each kind of coin.

Solution

Let \ (DP [i] \) represent the minimum number of components of \ (amount = I \). In case of transfer, it is obvious to enumerate every coin less than \ (amount \):

class Solution {
private:
    int dp[10004];
    
public:
    int coinChange(vector<int>& coins, int amount) {
        if(amount==0)return 0;
        int n = coins.size();
        int MAX = 9999999;
        for(int i=0;i<=amount;i++)dp[i] = MAX;
        for(int j=0;j<n;j++){
            if(amount>=coins[j])dp[coins[j]]=1;
        }
        
        for(int i=1;i<=amount;i++){
            for(int j=0;j<n;j++){
                if(coins[j]<=i){
                    dp[i] = min(dp[i], 1 + dp[i-coins[j]] );
                }
            }
        }
        if(dp[amount]>amount)return -1;
        else return dp[amount];
    }
};