LeetCode-Alg-860-Lemonade-Change

LeetCode-Alg-860-Lemonade-Change

Posted by Jae on June 19, 2019

1、题目

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 either 5, 10, or 20.

2、思路

该题是找零问题,可以使用贪心来求解,每次找零时都先从大面值的开始找,这样用掉的钞票数量最少,能更多的满足顾客。首先对于这道题需要记录收到的钞票数量 然后在每次处理顾客找零时,都从手里已有的大面值开始找。

3、实现

public class Solution {
    public boolean lemonadeChange(int[] bills) {
        // 存放已收到钞票数量
        int[] changeArr = new int[21];
        // 钞票面值
        int[] faceValue = {20, 10, 5};

        for (int bill : bills){
            int change = bill - 5;
            changeArr[bill]++;
            if (change > 0){
                for (int v : faceValue){
                    if (change >= v && changeArr[v] > 0){
                        int count = change / v;
                        if (changeArr[v] < count){
                            return false;
                        }else{
                            changeArr[v]-=count;
                            change = change - v * count;
                        }
                    }
                }
                if (change != 0){
                    return false;
                }
            }
        }
        return true;
    }
}