[CareerCup] 9.8 Represent N Cents 组成N分钱-优快云博客

本文提供了一种使用动态规划计算给定金额下不同硬币组合总数的方法。通过Java、Python和C++等语言实现了该算法，并给出了详细的代码示例。

9.8 Given an infinite number of quarters (25 cents), dimes (10 cents), nickels (5 cents) and pennies (1 cent), write code to calculate the number of ways of representing n cents.

给定一个钱数，用quarter，dime，nickle和penny来表示的方法总和。

Java:

public class Solution {
    /**
     * @param n an integer
     * @return an integer
     */
    public int waysNCents(int n) {
        int[] f = new int[n+1];
        f[0] = 1;
        int[] cents = new int[]{1, 5, 10, 25};
        for (int i = 0; i < 4; i++) 
            for (int j = 1; j <= n; j++) {
                if (j >= cents[i]) {
                    f[j] += f[j-cents[i]];
                }
            }
        return f[n];
    }
}

Python:

class Solution:
    # @param {int} n an integer
    # @return {int} an integer
    def waysNCents(self, n):
        # Write your code here
        cents = [1, 5, 10, 25]
        ways = [0 for _ in xrange(n + 1)]
        
        ways[0] = 1
        for cent in cents:
            for j in xrange(cent, n + 1):
                ways[j] += ways[j - cent]
        
        return ways[n]

C++:

class Solution {
public:
    /**
     * @param n an integer
     * @return an integer
     */
    int waysNCents(int n) {
        // Write your code here
        vector<int> cents = {1, 5, 10, 25};
        vector<int> ways(n + 1);

        ways[0] = 1;
        for (int i = 0; i < 4; ++i)
            for (int j = cents[i]; j <= n; ++j)
                ways[j] += ways[j - cents[i]];

        return ways[n];
    }
};

C++:

class Solution {
public:
    int makeChange(int n) {
        vector<int> denoms = {25, 10, 5, 1};
        vector<vector<int> > m(n + 1, vector<int>(denoms.size()));
        return makeChange(n, denoms, 0, m);
    }
    int makeChange(int amount, vector<int> denoms, int idx, vector<vector<int> > &m) {
        if (m[amount][idx] > 0) return m[amount][idx];
        if (idx >= denoms.size() - 1) return 1;
        int val = denoms[idx], res = 0;
        for (int i = 0; i * val <= amount; ++i) {
            int rem = amount - i * val;
            res += makeChange(rem, denoms, idx + 1, m);
        }
        m[amount][idx] = res;
        return res;
    }
};

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[LeetCode] 322. Coin Change 硬币找零