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  • leetcode 860. Lemonade Change

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