算法练习11：leetcode习题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:
Input: coins = [1, 2, 5], amount = 11
Output: 3
Explanation: 11 = 5 + 5 + 1

Example 2:
Input: coins = [2], amount = 3
Output: -1

Note:

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

算法思路

本周继续学习动态规划的问题，本题给出面额不同数量不限的几种硬币，要求总面额达到规定数值T的最少硬币数N。

动态规划里最重要的是找到合适的子问题，根据以往的经验，考虑原问题的前缀是个比较常见的思路，这道题也适合这个思路。
我们要求F(S)时，考虑先求出F(i),0<=i<S.

分解子问题的状态方程
F( i ) = min(F(i - C( j )),j=1:size( C ))， F(0) = 0
F( i )指的是总面值为i的最小硬币数。
C( j )表示第j种硬币的面值。
size( C )表示硬币种数。

C++代码

class Solution {
public:
    int coinChange(vector<int>& coins, int amount) {
        int coinnums[amount+1] = {0};
        for (int i = 1; i <= amount; ++i)
        {
            coinnums[i] = amount + 1;
        }

        for (int i = 1; i <= amount; ++i)
        {
            for (int j = 0; j < coins.size(); ++j)
            {
                if (coins[j] <= i && coinnums[i - coins[j]] + 1 < coinnums[i])
                {
                    coinnums[i] = coinnums[i - coins[j]] + 1;
                }
            }
        }

        return coinnums[amount] > amount ? -1 : coinnums[amount];
    }
};