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 either5
,10
, or20
.
贪心 时间复杂度O(n)
public boolean lemonadeChange(int[] bills) {//贪心 my int[] count=new int[3];//0为5,1为10,2为20,可用map代替 for (int i = 0; i < bills.length; i++) { if(5==bills[i]){ count[0]++; } else if(10==bills[i]){ count[1]++; if(0>=count[0]){ return false; } count[0]--; } else{ count[2]++; if((0>=count[1]&&3>count[0])||(0>=count[0])){ return false; } if(0>=count[1]){ count[0]-=3; } else{ count[1]--; count[0]--; } } } return true; }
简洁版
public boolean lemonadeChange(int[] bills) { int five = 0, ten = 0; for (int i : bills) { if (i == 5) five++; else if (i == 10) {five--; ten++;} else if (ten > 0) {ten--; five--;} else five -= 3; if (five < 0) return false; } return true; }