###给定硬币面额和种类,给定总金额,问有多少种组合方法; https://leetcode-cn.com/problems/coin-change-2/
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 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
硬币金额=重量,背包=金额,dp记录组合数
这题注意点就是说,一般要存满背包(这里的是金额),我们一般要用MAX去初始化数组,但是这题不是
* 这里用0初始化可以直接当做0种方法
####* 然后注意对于金额0,应该是有1种方法(就是什么都不装)(这里就能够让面额2硬币,对于金额2来说有d[2] = d[2]+d[2-2] = 1) ####* 可以重复使用,内层递增循环 ####* 另外注意更新的时候 状态转移方程应该是d[v] = d[v] + d[v-coins[i]],d[v]是上一行的结果,d[v-coins[i]]是这一行左边的结果 如图
class Solution {
public:
int change(int amount, vector<int>& coins) {
int dp[amount+1];
fill(dp,dp+amount+1,0);
dp[0] = 1;
int n = coins.size();
for(int i=0;i<n;i++){
for(int j=coins[i];j<=amount;j++){
dp[j] = dp[j]+dp[j-coins[i]];
}
}
return dp[amount];
}
};