LeetCode 322: Coin Change (硬币找零, DP经典题)

纸上得来终觉浅绝知此事要躬行2023-08-21 16:26

Coin Change
Medium
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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.length <= 12

1 <= coins[i] <= 231 - 1

0 <= amount <= 104

解法1：动态规划。注意此题不可用贪婪法。比如coins = {1, 3, 4}, amount = 6. 用贪婪法得到{4, 1, 1}这种组合，但实际上最优解是{3,3}。

cpp 复制代码

class Solution {
public:
    int coinChange(vector<int>& coins, int amount) {
        int n = coins.size();
        if (n == 0 || amount == 0) return 0;
        vector<int> dp(amount + 1, INT_MAX);
        for (int i = 1; i <= amount; i++) {
            for (auto j : coins) {
                if (i == j) dp[i] = 1;
                else if (i > j && dp[i - j] != INT_MAX) {
                    dp[i] = min(dp[i], dp[i - j] + 1);
                }
            }
        }
        if (dp[amount] == INT_MAX) return -1;
        return dp[amount];
    }
};