标签：not for ecif rect input Plan 心情 turn lse
https://leetcode.com/problems/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.
Note:
0 <= bills.length <= 10000bills[i] will be either 5, 10, or 20.代码：
class Solution {
public:
bool lemonadeChange(vector<int>& bills) {
int n = bills.size();
int n1 = 0, n2 = 0;
for(int i = 0; i < n; i ++) {
if(bills[i] == 5) n1 ++;
else if(bills[i] == 10) n1 --, n2 ++;
else if(n2 > 0) n2 --, n1 --;
else n1 -= 3;
if(n1 < 0) return false;
}
return true;
}
};
看下午的比赛有贪心 比赛之前就多找找贪心写一下吧 可能心情不是很好也会影响写的代码吧 脑子不是很在线 FH 应该还在吃午饭 吃完记得睡一会吧 盖好被子不要着凉哦 可以来一个柠檬茶提提神
#Leetcode# 860. Lemonade Change
标签：not for ecif rect input Plan 心情 turn lse
原文地址：https://www.cnblogs.com/zlrrrr/p/10303114.html
