LeetCode 518. 零钱兑换 II 中等
我的解题
是参考 labuladong 经典动态规划:完全背包问题 中的解法。
#include <bits/stdc++.h>
using namespace std;
class Solution
{
public:
int change(int amount, vector<int> &coins)
{
int n = coins.size();
vector<vector<int>> dp(n + 1, vector<int>(amount + 1, 0));
for (int i = 0; i <= n; ++i)
{
dp[i][0] = 1;
}
for (int i = 1; i <= n; ++i)
for (int j = 1; j <= amount; ++j)
{
if (j - coins[i - 1] >= 0) // 能够装下第i个物品
{
dp[i][j] = dp[i - 1][j] + dp[i][j - coins[i - 1]];
}
else // 装不下第i个物品
{
dp[i][j] = dp[i - 1][j];
}
}
return dp[n][amount];
}
};
ostream& operator<<(ostream &stream, vector<int> v)
{
for (auto &&i : v)
stream << i;
return stream;
}
// Driver code
int main()
{
Solution s;
return 0;
}
// g++ test.cpp --std=c++11 -pedantic -Wall -Wextra