Coin

Coin"/>

Coin

题目描述：

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.
 (你有不同面额的硬币和一个总金额。
  写一个函数来计算你所需要的最少的硬币数。
  如果这笔钱不能由硬币的任何组合构成，则返回-1。)
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.
（每种硬币你有无穷个。）

思路：动态规划题目，模型为完全背包，以dp[i]来表示当价值为i时所要使用的硬币数，那么dp[i-coins[j]]+1就是当价值为i-coins[j]时再加一枚coins[j]就可以到达价值i，通过这种思想就可以得到递推式dp[i]=Math.min(dp[i],dp[i-coins[j]]+1)，当价值为i时无法形成一个硬币组合时dp[i]就为初始赋予给其的值。

public class Coin_Change {public static int coinChange(int[] coins, int amount) {if(coins == null||coins.length == 0)return 0;Arrays.sort(coins);int dp[] = new int[amount+1];//当价格为0时没有解dp[0] = 0;//int类型最大值正数为2147483647，加1后溢出的结果 -2147483648//防止后面在递推式中发生2147483647+1for(int i=1;i<=amount;i++){dp[i]=Integer.MAX_VALUE-1;}for(int i=1;i<=amount;i++){for(int j=0;j<coins.length;j++){if(i>=coins[j])dp[i] = Math.min(dp[i], dp[i-coins[j]]+1);}}if(dp[amount] == Integer.MAX_VALUE-1)return -1;return dp[amount];}public static void main(String[] args) {int coins[] = {3,2,4};int amount = 21;System.out.println(coinChange(coins,amount));}
}