LeetCode 322. 零钱兑换（DP）

示例 1:
输入: coins = [1, 2, 5], amount = 11
输出: 3 
解释: 11 = 5 + 5 + 1

示例 2:
输入: coins = [2], amount = 3
输出: -1

说明:
你可以认为每种硬币的数量是无限的。

class Solution//回溯
{
public:
    int coinChange(vector<int>& coins, int amount)
    {
        int minPieces = INT_MAX;
        money(coins, amount, 0, 0, minPieces);
        if(minPieces != INT_MAX)
            return minPieces;
        else
            return -1;
    }
    void money(vector<int>& coins, int &amount, long curMoney, int curPieces, int &minPieces)
    {
        if(curMoney > amount)
        {
            curPieces = INT_MAX;//表示不存在
            return;
        }
        if(curMoney == amount)
        {
            if(curPieces < minPieces)
                minPieces = curPieces;
            return;
        }
        for(int i = 0; i < coins.size(); ++i)
        {
            money(coins, amount,curMoney+coins[i], curPieces+1, minPieces)；
        }
    }
};

class Solution {
public:
    int coinChange(vector<int>& coins, int amount) {
        if(amount < 1)
            return 0;
        int states[amount+1];
        int i, j;
        for(i = 0; i <= amount; ++i)
            states[i] = amount+1;//找i元钱，不可能需要amount+1张
        states[0] = 0;
        for(i = 0; i <= amount; ++i)
        {
            for(j = 0; j < coins.size(); ++j)
            {
                if(i-coins[j] >= 0 && states[i] > states[i-coins[j]]+1)
                {
                    states[i] = states[i-coins[j]]+1;
                }
            }
        }
        if(states[amount] != amount+1)
            return states[amount];
        else
            return -1;
    }
};

class Solution { //2020.3.8
public:
    int coinChange(vector<int>& coins, int amount) {
        if(amount <= 0)
            return 0;
        int n = coins.size(), j, m;
        vector<int> dp(amount+1, INT_MAX);
        dp[0] = 0;
        for(j = 0; j <= amount; ++j)//遍历所有的金额数
        {
            if(dp[j] != INT_MAX)
            {
                for(m = 0; m < n; ++m)
                {
                    if(j <= amount-coins[m])
                        dp[j+coins[m]] = min(dp[j]+1, dp[j+coins[m]]);
                }
            }
        }
        if(dp[amount] != INT_MAX)
            return dp[amount];
        return -1;
    }
};