[LeetCode] (medium) 322. Coin Change

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本文探讨了硬币找零问题，通过动态规划算法计算组成特定金额所需的最少硬币数量。介绍了如何使用C++实现该算法，并通过具体实例展示了其运行过程和优化技巧。

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https://leetcode.com/problems/coin-change/

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.

dp，但由一些很奇怪的优化点，很细微，但影响谜之很大，尤其是最后一句return，都不在循环里，也不涉及空间分配的问题，竟然能影响4ms

class Solution {
public:
    int coinChange(vector<int>& coins, int amount) {
        // vector<int> dp(amount+1);
        int dp[amount+1];   //48ms -> 40ms
        dp[0] = 0;
        int tem;
        for(int i = 1; i <= amount; ++i){
            dp[i] = amount+1;
            for(int j = 0; j < coins.size(); ++j){
                if(i < coins[j]) continue;
                tem = dp[i-coins[j]]+1;     //68ms -> 48ms
                if(tem < dp[i]) dp[i] = tem;
                // dp[i] = min(dp[i], dp[i-coins[j]]+1);
            }
        }
        return dp[amount] > amount ? -1 : dp[amount];   //40ms -> 36ms
        // return dp[amount] == amount+1 ? -1 : dp[amount];
    }
};