leetcode 860. Lemonade Change
阿新 • • 發佈:2018-07-01
cau custom tor input from Go ive ecif ransac
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:貪心,優先使用10元的去找零錢。
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