卖萌的弱渣

I am stupid, I am hungry.

Coin Change

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.

Example

  • Example 1: coins = [1, 2, 5], amount = 11 return 3 (11 = 5 + 5 + 1)

  • Example 2: coins = [2], amount = 3 return -1.

Note

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

Solution

  • Java
(Coin-Change.java) download
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public class Solution {
    public int coinChange(int[] coins, int amount) {
        if (coins == null || amount <= 0 || coins.length == 0)
            return 0;

        int[] minNumber = new int[amount+1];
        for (int i=1;i<=amount;i++){
            // invalid
            minNumber[i] = Integer.MAX_VALUE;
            for (int j=0;j<coins.length;j++){
                if(coins[j] <= i && minNumber[i-coins[j]] != Integer.MAX_VALUE)
                    minNumber[i] = Integer.min(minNumber[i], 1+minNumber[i-coins[j]]);

            }
        }
        if (minNumber[amount]==Integer.MAX_VALUE)
            return -1;
        else
            return minNumber[amount];

    }
}
  • Python
(Coin-Change.py) download
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class Solution(object):
    def coinChange(self, coins, amount):
        """
        :type coins: List[int]
        :type amount: int
        :rtype: int
        """
        # dp[i]: the least step needed to reach i dollars
        # dp[i+coin[j]] = min(dp[i]+1,dp[i+coin[j]])
        # dp[0] = 0
        dp = [0]+[0x7ffffffe] * (amount)

        for i in range(amount+1):
            for j in coins:
                if i +j <=amount:
                    dp[i+j] = min(dp[i]+1,dp[i+j])

         return dp[amount] if dp[amount] != 0x7ffffffe else -1