动态规划解决最少硬币凑成m元钱

322. Coin Change

 

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

Example 2:
coins = [2], amount = 3

return -1.

public class Solution{    /*长度为m的数组c[1...m]中存放一系列子结果，即c[i]为要凑的钱数为i时所需的最少硬币数，则c[m]为所求  当要找的钱数i(1<i<m)与当前所试探的硬币面值k相等时，结果为1，即c[i]=1  当i大于当前所试探硬币面值k时，若c[i]为0，即还未赋过值，且c[i-k]不为0，即从i元钱中刨去k元后剩下的钱数可以找开，             则c[i]=c[i-k]+1        若c[i]不为0，即已赋过值，则c[i]为c[i-k]+1和c[i]中较小的*/    public int coinChange(int[] coins, int amount)     {        for(int i=0;i<coins.length;i++)        {            System.out.println(coins[i]);        }        int m=amount;        int n=coins.length;        int[] a=coins;        int[] c=new int[m+1];//fu chu zhi 0        if(amount==0)            return 0;               for(int i=0;i<n;i++)        {            if(a[i]<=m)            {                c[a[i]]=1;            }                }        for(int i=1;i<=m;i++)        {            for(int j=0;j<n;j++)            {                if(i>a[j])                {                    if(c[i]==0&&c[i-a[j]]!=0)                    {                        c[i]=c[i-a[j]]+1;                        System.out.println("c"+i+":"+c[i]);                    }                     else                    {                        if(c[i-a[j]]!=0)                        {                            c[i]=c[i-a[j]]+1<c[i]?c[i-a[j]]+1:c[i];                            System.out.println("c"+i+":"+c[i]);                        }                    }                }                           }        }        if(c[m]==0)        return -1;        else        return c[m];            }}

另一种更有效的方法

public int coinChange(int[] coins, int amount) {    if (amount < 1) return 0;    int[] dp = new int[amount + 1];     Arrays.fill(dp, Integer.MAX_VALUE);    dp[0] = 0;    for (int coin : coins) {        for (int i = coin; i <= amount; i++) {            if (dp[i - coin] != Integer.MAX_VALUE) {                dp[i] = Math.min(dp[i], dp[i - coin] + 1);            }        }    }    return dp[amount] == Integer.MAX_VALUE ? -1 : dp[amount];}