LINTCODE—— Coin Change II

LINTCODE—— Coin Change II

题目：给你一个amount,再给你一个硬币数组，coins，数组的值表示为硬币的价值，硬币可以无限使用，求从数组中选择一定的组合满足硬币总价值等于amount有多少种情况

思路：动态规划，类似于无限背包问题，假定dp[j]为从数组coins中选取总价值为j的情况，则不难得出有：dp[j] = dp[j] +dp[j-coins[i]],coins[i]为第i个硬币的价值

class Solution {public:    /**     * @param amount: a total amount of money amount     * @param coins: the denomination of each coin     * @return: the number of combinations that make up the amount     */    int change(int amount, vector<int> &coins) {        // write your code here        // write your code here        // 无限背包问题变种        // 初始化dp数组        vector<int> dp(amount+1,0);        dp[0] = 1;        for (int i = 0 ; i < coins.size(); i++) {            for (int j = 1 ; j <= amount; j++) {                if(j >= coins[i])                    dp[j] += dp[j-coins[i]];            }        }        return dp[amount];       }};