322. Coin Change
- Hardness: \(\color{orange}\textsf{Medium}\)
- Ralated Topics:
Array、Dynamic Programming、Breadth-First Search
一、題目
You are given an integer array coins representing coins of different denominations and an integer amount representing a total amount of money.
Return 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.
You may assume that you have an infinite number of each kind of coin.
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
Example 3:
- Input: coins = [1], amount = 0
- Output: 0
Constraints:
1 <= coins <= 121 <= coins[i] <= 2^31 - 10 <= amount <= 10^4
二、分析
- 這一題動態規劃是不定序列型框架的題型：
- 令
dp[n]為組成n有幾種可能性。 dp[n] = sum(dp[n-nums[i]])- 以題目 coins = [1,2,5] 為例，
dp[11]可以為dp[11-1] + dp[11-2] + dp[11-5]。
- 以題目 coins = [1,2,5] 為例，
三、解題
1. Dynamic Programming
- Time complexity: \(O(n)\)
- Space complexity: \(O(n)\)
int coinChange(vector<int>& coins, int amount) {
vector<int> dp(amount+1, 20000);
dp[0] = 0;
for (int i = 0; i < coins.size(); i++){
for (int j = coins[i]; j <= amount; j++){
dp[j] = min(dp[j], dp[j-coins[i]] + 1);
}
}
return dp[amount] >= 20000 ? -1 : dp[amount];
}