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:
Input: coins =[1, 2, 5], amount =11Output:3Explanation: 11 = 5 + 5 + 1
Example 2:
Input: coins =[2], amount =3Output: -1
题目大意：
使用几种硬币找零钱，每种硬币可多次使用，求所需硬币最少数目。若硬币之和不能等于零钱总数，输出-1.
解题思路：
用递归的动态规划解决
1 class Solution { 2 public: 3 4 const int inf = 1000000000; 5 vector<vector<int>> v; 6 int search(vector<int>& coins, int amount, int idx) { 7 if (amount == 0) 8 return 0; 9 if (idx >= coins.size() || amount < 0) 10 return inf; 11 if (v[idx][amount] >= 0) 12 return v[idx][amount]; 13 int a = search(coins, amount - coins[idx], idx); 14 int b = search(coins, amount, idx + 1); 15 v[idx][amount] = min(a + 1, b); 16 return v[idx][amount]; 17 } 18 19 int coinChange(vector<int>& coins, int amount) { 20 v.resize(coins.size(), vector<int>(amount + 1, -1)); 21 int ans = search(coins, amount, 0); 22 if (ans < inf) 23 return ans; 24 return -1; 25 } 26 };
原文：https://www.cnblogs.com/lxc1910/p/10493495.html