[LeetCode] 860. Lemonade Change 買檸檬找零

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.

這道題說是有不少檸檬，每一個賣5刀，顧客可能會提供5刀，10刀，20刀的鈔票，咱們剛開始的時候並無零錢，只有收到顧客的5刀，或者 10 刀能夠用來給顧客找錢，固然若是第一個顧客就給 10 刀或者 20 刀，那麼是沒法找零的，這裏就問最終是否可以都成功找零。博主最早想到的方法是首先用一個 HashMap 來分別統計出5刀，10 刀，和 20 刀鈔票的個數，而後再來統一分析是否能成功找零。因爲 10 刀的鈔票須要5刀的找零，20 刀的鈔票能夠用1張 10 刀和1張5刀，或者3張5刀的鈔票，因此至少須要1張5刀，那麼當前5刀的個數必定不能小於 10 刀和 20 刀的個數以後，不然沒法成功找零。因爲 20 刀能夠用 10 刀來找零，每一個 10 刀能夠節省兩個5刀，可是爲了得到每張 10 刀，咱們還得付出一張5刀的找零，因此實際上用 10 刀來找零隻能省下1張5刀鈔票，可是假如 10 刀的個數不夠，那麼每張 20 刀的鈔票仍是須要3張5刀的鈔票來找零的，因此判斷若5刀鈔票的個數小於 20 刀鈔票個數的三倍減去 10 刀鈔票的個數，直接返回 false。for 循環退出後返回 true，參見代碼以下：

解法一：

class Solution {
public:
    bool lemonadeChange(vector<int>& bills) {
        unordered_map<int, int> cnt;
        for (int bill : bills) {
            ++cnt[bill];
            if (cnt[5] < cnt[20] + cnt[10]) return false;
            if (cnt[5] < 3 * cnt[20] - cnt[10]) return false;
        }
        return true;
    }
};

實際上咱們並不須要一直保留全部鈔票的個數，當某些鈔票被看成零錢給了，就沒有必要繼續留着它們的個數了。其實上咱們只關心當前還剩餘的5刀和 10 刀鈔票的個數，用兩個變量 five 和 ten 來記錄。而後遍歷全部的鈔票，假如遇到5刀鈔票，則 five 自增1，若遇到 10 刀鈔票，則須要找零5刀，則 five 自減1，ten 自增1。不然遇到的就是 20 刀的了，若還有 10 刀的鈔票話，就先用 10 刀找零，則 ten 自減1，再用一張5刀找零，five 自減1。若沒有 10 刀了，則用三張5刀找零，five 自減3。找零了後檢測若此時5刀鈔票個數爲負數了，則直接返回 false，參見代碼以下：

解法二：

class Solution {
public:
    bool lemonadeChange(vector<int>& bills) {
        int five = 0, ten = 0;
        for (int bill : bills) {
            if (bill == 5) ++five;
            else if (bill == 10) { --five; ++ten; }
            else if (ten > 0) { --ten; --five; }
            else five -= 3;
            if (five < 0) return false;
        }
        return true;
    }
};

Github 同步地址:

https://github.com/grandyang/leetcode/issues/860

參考資料：

https://leetcode.com/problems/lemonade-change/

https://leetcode.com/problems/lemonade-change/discuss/143719/C%2B%2BJavaPython-Straight-Forward

LeetCode All in One 題目講解彙總(持續更新中...)