动态规划解322. Coin Change

题目

You are given coins of different denominations and a total amount of money amount. Write a function to compute the fewest number of coins that you need to make up that amount. If that amount of money cannot be made up by any combination of the coins, return -1.

Example 1:coins = [1, 2, 5], amount = 11return 3 (11 = 5 + 5 + 1)

Example 2:coins = [2], amount = 3return -1.

Note:You may assume that you have an infinite number of each kind of coin.

题意分析

题目意思是说，给一系列的硬币，求这些硬币最小组合成目标数，如果能组合成目标数，则返回最小组合硬币的个数，如果不能组合成目标数，则返回-1

思路分析

这是一道典型的动态规划，要求目标数的最小硬币组合数，可以通过将其减去每种硬币的值（如果目标数比某种硬币值大的话就进行该操作），然后取这些值中的最小值，那么就可以得到答案。为了可以辨别出是否能组合成目标数，可以将每个初始值设为目标数+1，如果最后没有改变初始值，则说明不能组合成该目标数

AC代码

class Solution {public:    int coinChange(vector<int>& coins, int amount) {        vector<int> vec;        vec.push_back(0);        for (int i = 1; i <= amount; ++i) {            vec.push_back(amount+1);        }        for (int i = 1; i <= amount; ++i) {            for (int j = 0; j < coins.size(); ++j) {                if (i >= coins[j]) {                    vec[i] = min(vec[i], vec[i-coins[j]] + 1);                }            }        }        if (vec[amount] == amount + 1) {            return -1;        } else {            return vec[amount];        }    }};