# 518. Coin Change 2

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
``````

dp[i][j] = dp[i - 1][j] + (j >= coins[i - 1] ? dp[i][j - coins[i - 1]] : 0)

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

``````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];
}
};
``````

``````class Solution {
public:
int change(int amount, vector<int>& coins) {
if (amount == 0) return 1;
if (coins.empty()) return 0;
map<pair<int, int>, int> memo;
return helper(amount, coins, 0, memo);
}
int helper(int amount, vector<int>& coins, int idx, map<pair<int, int>, int>& memo) {
if (amount == 0) return 1;
else if (idx >= coins.size()) return 0;
else if (idx == coins.size() - 1) return amount % coins[idx] == 0;
if (memo.count({amount, idx})) return memo[{amount, idx}];
int val = coins[idx], res = 0;
for (int i = 0; i * val <= amount; ++i) {
int rem = amount - i * val;
res += helper(rem, coins, idx + 1, memo);
}
return memo[{amount, idx}] = res;
}
};
``````

Github 同步地址：

https://github.com/grandyang/leetcode/issues/518

Coin Change

9.8 Represent N Cents 美分的组成

https://leetcode.com/problems/coin-change-2/

https://leetcode.com/problems/coin-change-2/discuss/141076/Logical-Thinking-with-Clear-Java-Code

https://leetcode.com/problems/coin-change-2/discuss/99212/Knapsack-problem-Java-solution-with-thinking-process-O(nm)-Time-and-O(m)-Space

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