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  • [CareerCup] 9.8 Represent N Cents 组成N分钱

    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;
        }
    };
    

      

    类似题目:

    [LeetCode] 322. Coin Change 硬币找零

    CareerCup Questions List 职业杯题目列表

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  • 原文地址:https://www.cnblogs.com/lightwindy/p/8674187.html
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