【数据结构】算法 Lemonade Change 柠檬水零钱

2021/4/10 12:27:58

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目录
  • Lemonade Change 任务调度器
    • 思路
    • tag

Lemonade Change 任务调度器

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.

一杯柠檬水5$,开始时没有任何钱,每个客户一次买一杯,但是可能会支付5,10,20的现金,如果现有的钱能够完成所有客户的找零,最后返回true,否则false

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.

思路

遇到需要找零的情况,先使用10+5,不够再使用5*3,原则是保留更多的5以便后面找零。

class Solution {
    public boolean lemonadeChange(int[] bills) {
int cnt5= 0,cnt10 =0;
        for (int bill : bills) {
            switch (bill){
                case 5:
                    cnt5+=1;
                    break;
                case 10:
                    if (cnt5==0){
                        return false;
                    }
                    cnt5-=1;
                    cnt10+=1;
                    break;
                case 20:
                    if(cnt5>=1&&cnt10>=1){
                        cnt5-=1;
                        cnt10-=1;
                       
                    }
                    else if (cnt5>=3){
                        cnt5 = cnt5-3;
                        
                    }
                    else {
                        return false;
                    }
                    break;
            }
        }
        return true;
    }
}

tag

贪心



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