1546. 零钱问题

描述

小明是一个销售员，客人在他的地方买了东西，付给了小明一定面值的钱之后，小明需要把多余的钱退给客人。客人付给了小明n，小明的东西的售价为m，小明能退回给客人的面额只能为[100,50,20,10,5,2,1]的组合。现在小明想要使纸币数量之和最小，请返回这个最小值。

1 <= m <= n <= 10000000001≤m≤n≤1000000000

您在真实的面试中是否遇到过这个题？
样例
Give n=100, m=29, return 3.

100-29=71
Ming retrieved one 50, one 20 and one 1.
So the answer is 3.

Give n=50, m=5, return 3.

50-5=45
Ming retrieved two 20 and one 5.
So the answer is 3.

public class Solution {
    /**
     * @param n: The guest paid
     * @param m: the price
     * @return: the sum of the number of banknotes
     */
    public int coinProblem(int n, int m) {
        // Write your code here
        int mon = n - m;
        
        int[] comb = {100, 50, 20, 10, 5, 2, 1};
        
        int res = 0;
        for (int i=0; i<comb.length; i++){
            res += mon / comb[i];
            mon = mon % comb[i];
            
            
        } 
        return res;
    }
}