LeetCode-860. 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 <= 10000
  • bills[i] will be either 510, or 20.

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;
    }

 

posted @ 2019-03-25 21:36  月半榨菜  阅读(68)  评论(0编辑  收藏  举报