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.

一刷
題解：
方法1: 用dynamic programming, int[] dp保存0-amout的結(jié)果。
但是這樣會造成array out of boundary, 原因是空間復(fù)雜度可以達(dá)到O(amount)

public static int coinChange(int[] coins, int amount) {
        if(amount==0) return 0;
     
        int[] dp = new int [amount+1];
        dp[0]=0; // do not need any coin to get 0 amount
        for(int i=1;i<=amount; i++)
            dp[i]= Integer.MAX_VALUE;
     
        for(int i=0; i<=amount; i++){
            for(int coin: coins){
                if(i+coin <=amount){
                    if(dp[i]!=Integer.MAX_VALUE){
                        dp[i+coin] = Math.min(dp[i+coin], dp[i]+1);
                    }
                }
            }
        }
     
        if(dp[amount] >= Integer.MAX_VALUE)
            return -1;
     
        return dp[amount];
    }

方法2: Breath First Search (BFS)
amountQueue存儲當(dāng)前的coin sum up的值，step存儲對應(yīng)的硬幣數(shù)目。
但是會出現(xiàn)超時的問題

public static int coinChange(int[] coins, int amount) {
        if(amount == 0) return 0;
        
        LinkedList<Integer> amountQueue = new LinkedList<Integer>();
        LinkedList<Integer> stepQueue = new LinkedList<Integer>();
        
        amountQueue.offer(0);//attach to the tail
        stepQueue.offer(0);
        
        while(amountQueue.size()>0){
            int temp = amountQueue.poll();
            int step = stepQueue.poll();
            
            if(temp == amount) return step;
            
            for(int coin : coins){
                if(temp<=amount){
                    if(!amountQueue.contains(temp+coin)){
                        amountQueue.offer(temp+coin);
                        stepQueue.offer(step+1);
                    }
                }
            }
        }
        
        return -1;
    }

二刷
用map替代dp數(shù)組做dp

class Solution {
    Map<Integer, Integer> amountDict = new HashMap<>();
    public int coinChange(int[] coins, int amount) {
        if(amount == 0) return 0;
        if(amountDict.containsKey(amount)) return amountDict.get(amount);
        int n = amount + 1;
        for(int coin : coins){
            int curr = 0;
            if(amount>=coin){
                int next = coinChange(coins, amount-coin);
                if(next>=0) curr = next+1;
            }
            if(curr>0) n = Math.min(n, curr);
        }
        
        int finalCount = (n==amount+1)? -1:n;
        amountDict.put(amount, finalCount);
        return finalCount;
    }
}