Lemonade Change

这题不难，核心是找零问题，首先看下题目：

At a lemonade stand, each lemonade costs $5. 

Customers are standing in a queue to buy from you, and order one at a time (in the order specified by bills).

Each customer will only buy one lemonade and pay with either a $5, $10, or $20 bill.  You must provide the correct change to each customer, so that the net transaction is that the customer pays $5.

Note that you don't have any change in hand at first.

Return true if and only if you can provide every customer with correct change.

 

Example 1:

Input: [5,5,5,10,20]
Output: true
Explanation: 
From the first 3 customers, we collect three $5 bills in order.
From the fourth customer, we collect a $10 bill and give back a $5.
From the fifth customer, we give a $10 bill and a $5 bill.
Since all customers got correct change, we output true.
Example 2:

Input: [5,5,10]
Output: true
Example 3:

Input: [10,10]
Output: false
Example 4:

Input: [5,5,10,10,20]
Output: false
Explanation: 
From the first two customers in order, we collect two $5 bills.
For the next two customers in order, we collect a $10 bill and give back a $5 bill.
For the last customer, we can't give change of $15 back because we only have two $10 bills.
Since not every customer received correct change, the answer is false.
 

Note:

0 <= bills.length <= 10000
bills[i] will be either 5, 10, or 20.

主要解题思路在于，找零的时候，用最大的零钱先去找开来最大能找的钱，剩下的钱用较小面值的零钱去找，代码如下：

static bool Main(const vector<int>& bills) {
        int nChanges[2] = {0};
        int nPoint[2] = {5, 10};
        int nRound = 0;
        for (auto v : bills) {
            if (v == 5) {
                // Just put it into wallet
                nChanges[0]++;
            } else {
                int nNeedChange = v - 5;

                for (int i = sizeof(nPoint) / sizeof(nPoint[0]) - 1; i >= 0 ; i--) {
                    int n1 = nNeedChange / nPoint[i];
                    int n2 = nChanges[i];
                    int n = n1 < n2 ? n1 : n2;
                    nNeedChange -= nPoint[i] * n;
                    nChanges[i] -= n;
                }

                if (0 != nNeedChange) {
                    return false;
                }
                if (v == 10) {
                    nChanges[1]++;
                }
            }
            nRound++;
        }
        return true;
    }