Change problem

1, in the lemonade stalls, each cup of lemonade for $ 5.

Customers line up to buy your product, (the bill in order to pay bills) to buy a drink.

Each customer only buy a cup of lemonade, then you pay $ 5, $ 10 or $ 20. You have the right to change for each customer, which means that net trade is each customer to pay you $ 5.

Note that, at the beginning you do not have any change.

If you give the correct change for each customer, returns true, otherwise false.

Example 1:

Enter: [5,5,5,10,20]
Output: true
explanation:
the top three customers there, we charge three five-dollar bill in order.
No. 4 customers there, we charge a ten dollar bill, and returned $ 5.
5th customer there, we also find a ten dollar bill and a five dollar bill.
Since all customers have been properly give change, so we output true.
Example 2:

Input: [5,5,10]
Output: true
Example 3:

Input: [10,10]
Output: false
Example 4:

Enter: [5,5,10,10,20]
output: false
explanation:
before two customers there, we charge two five-dollar bill in order.
For the next two customers, we charge a ten dollar bill, and then to return $ 5.
For the last customers, we can not return the $ 15, because we are now only two $ 10 bills.
Since not every customer get the right give change, so the answer is false.
 

prompt:

0 <= bills.length <= 10000
Bills [I] is not 10 or 20 5 

Source: stay button (LeetCode)
link: https: //leetcode-cn.com/problems/lemonade-change
copyrighted by deduction from all networks. Commercial reprint please contact the authorized official, non-commercial reprint please indicate the source.

 

 1 class Solution {
 2 public:
 3     bool lemonadeChange(vector<int>& bills) {
 4         int a = 0,b = 0;
 5         for(int i = 0;i<bills.size();i++){
 6             if(bills[i] == 5)
 7                 a++;
 8             if(bills[i] == 10){
 9                 if(a == 0)
10                     return false;
11                 a--;
12                 b++;
13             }
14             if(bills[i] == 20){
15                 if (a > 0 && b > 0){
16                     a--;
17                     b--;                    
18                 }
19                 else if(a >= 3)
20                     a -= 3;
21                 else
22                     return false;
23 
24             }
25         }
26         return true;
27 
28     }
29 };

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2 

Given the different denominations of coins coins and a total amount of amount. We can write a function to calculate the minimum number of coins needed to make up the total amount. If there is no combination can be composed of any of a total amount of coins, return -1.

Example 1:

Input: coins = [1, 2, 5], amount = 11
Output: 3
Explanation: 11 = 5 + 5 + 1
Example 2:

Input: coins = [2], amount = 3
Output: -1

Source: stay button (LeetCode)
link: https: //leetcode-cn.com/problems/coin-change
copyrighted by deduction from all networks. Commercial reprint please contact the authorized official, non-commercial reprint please indicate the source.

 

 1 #include<algorithm>
 2 class Solution {
 3 public:
 4     int coinChange(vector<int>& coins, int amount) {
 5     int max = amount + 1;
 6     int dp[amount+1];
 7     for(int i = 0;i<amount+1;i++)
 8         dp[i] = max;
 9     dp[0] = 0;
10     for(int i = 1;i <= amount;i++)
11         for(int j = 0;j<coins.size();j++){
12             if(coins[j] <= i)
13                 dp[i] = min(dp[i],dp[i-coins[j]]+1);
14         }
15     return dp[amount]>amount?-1:dp[amount];
16 }
17 };

View Code