算法设计Week16 LeetCode Algorithms Problem #322 Coin Change

题目描述：

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 = 11
return 3 (11 = 5 + 5 + 1)

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

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

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题目分析：

使用动态规划的方法，使用count[i]表示钱的总数为i时最少需要的硬币数目。显然，当i = 0时count[0] = 0。如果i比某一个coins的值大时（coins[j]表示），那么我们有：
count[i] = min(count[i - coins[j]] + 1, count[i]);

那么，当amount无法由coins组合出来时，要如何判断呢？我们可以在初始化count数组时，将初始值均赋为一个不应该出现的“最大”值，如amount + 1。这样，在返回值的时候进行一个简单的判断即可。

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代码：

算法的时间复杂度为O(n∗amount)，空间复杂度为O(amount)。

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