leetcode 860. Lemonade Change

标签：first   turn   code   pre   AC   and   mon   collect   nsa   

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.

思路：贪心，优先使用10元的去找零钱。

class Solution {
public:
    bool lemonadeChange(vector<int>& bills) {
        queue<int> q;
        int num5 = 0;
        int num10 = 0;
        for (int i = 0; i < bills.size(); ++i) {
            if (bills[i] == 5) {
                num5++;
            }
            else {
                int x = bills[i];
                if (bills[i] == 20) {
                    while (bills[i] > 10 && num10 > 0) {
                        bills[i] -= 10;
                        num10--;
                    }
                    while (bills[i] > 5 && num5 > 0) {
                        bills[i] -= 5;
                        num5--;
                    }
                    if (bills[i] > 5) {
                        return false;
                    }
                }
                else {
                    while (bills[i] > 5 && num5 > 0) {
                        bills[i] -= 5;
                        num5--;
                    }
                    if (bills[i] > 5) {
                        return false;
                    }
                    num10++;
                }
            } 
        }
        return true;
    }
};

leetcode 860. Lemonade Change

标签：first   turn   code   pre   AC   and   mon   collect   nsa   

原文地址：https://www.cnblogs.com/pk28/p/9250169.html