leetcode 518. Coin Change 2 类似背包问题 + 很简单的动态规划DP解决

You are given coins of different denominations and a total amount of money. Write a function to compute the number of combinations that make up that amount. You may assume that you have infinite number of each kind of coin.

Note: You can assume that

0 <= amount <= 5000
1 <= coin <= 5000
the number of coins is less than 500
the answer is guaranteed to fit into signed 32-bit integer
Example 1:

Input: amount = 5, coins = [1, 2, 5]
Output: 4
Explanation: there are four ways to make up the amount:
5=5
5=2+2+1
5=2+1+1+1
5=1+1+1+1+1
Example 2:

Input: amount = 3, coins = [2]
Output: 0
Explanation: the amount of 3 cannot be made up just with coins of 2.
Example 3:

Input: amount = 10, coins = [10]
Output: 1

本题题意很简单，最初想到的是DFS深度优先遍历的做法，后来看到了一个DP做法，

dp[i]表示组成钱数i的不同方法。其实最开始的时候，博主就想着用一维的dp数组来写，但是博主开始想的方法是把里面两个for循环调换了一个位置，结果计算的种类数要大于正确答案，所以一定要注意for循环的顺序不能搞反，

建议和这一道题leetcode 322. Coin Change 类似背包问题 一起学习

还有这一道题leetcode 377. Combination Sum IV 组合之和 + DP动态规划 + DFS深度优先遍历一起学习

这两到的DP的出发点是不一样的，322是尽量凑够amout，然后求解最小的数量，本题则是求解能够组成amount的所有的情况的计数，所以这两道题很类似，但是不一样

还建议和leetcode 279. Perfect Squares 类似背包问题 + 很简单的动态规划DP解决 一起学习

还有leetcode 474. Ones and Zeroes若干0和1组成字符串最大数量+动态规划DP+背包问题

代码如下：

#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 

using namespace std;

class Solution 
{
public:
    int change(int amount, vector<int>& coins) 
    {
        vector<int> dp(amount + 1, 0);
        dp[0] = 1;
        for (int coin : coins) 
        {
            for (int i = coin; i <= amount; ++i) 
            {
                dp[i] += dp[i - coin];
            }
        }
        return dp[amount];
    }
};