dp:322. Coin Change 自下而上的dp

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.

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

 

Example 1:

Input: coins = [1,2,5], amount = 11
Output: 3
Explanation: 11 = 5 + 5 + 1

Example 2:

Input: coins = [2], amount = 3
Output: -1

Example 3:

Input: coins = [1], amount = 0
Output: 0

Example 4:

Input: coins = [1], amount = 1
Output: 1

Example 5:

Input: coins = [1], amount = 2
Output: 2

 

Constraints:

• 1 <= coins.length <= 12
• 1 <= coins[i] <= 231 - 1
• 0 <= amount <= 104

class Solution {
public:
    //无限背包问题
    int coinChange(vector<int>& coins, int amount) {
        vector<int> dp(amount+1,amount+1);
        dp[0] = 0;
        for(int i=1;i<= amount;i++){
            for(int j=0;j<coins.size();j++){
                //dp[i]初值都是amount+1
                if(i>=coins[j]) dp[i] = min(dp[i-coins[j]]+1,dp[i]);
            }
        }
        return dp[amount] < amount+1 ? dp[amount]:-1;
    }
};