LeetCode-322. Coin Change [C++][Java]

LeetCode-322. Coin Change[这里是图片001]https://leetcode.com/problems/coin-change/

You are given an integer arraycoinsrepresenting coins of different denominations and an integeramountrepresenting 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 <= 121 <= coins[i] <= 2^31?- 10 <= amount <= 10^4 解题思路

完全背包问题。

class Solution {
public:
    int coinChange(vector& coins, int amount) {
        if (coins.empty()) {return -1;}
        vector dp(amount + 1, amount + 2);
        dp[0] = 0;
        for (auto c : coins) {
            for (auto j = c; j <= amount; ++j) {
                dp[j] = min(dp[j], 1 + dp[j-c]);
            }
        }
        return dp[amount] == amount + 2 ? -1 : dp[amount];
    }
};

class Solution {
    public int coinChange(int[] coins, int amount) {
        if (coins.length == 0) {return -1;}
        int[] dp = new int[amount + 1];
        Arrays.fill(dp, amount + 2);
        dp[0] = 0;
        for (int c : coins) {
            for (int j = c; j <= amount; ++j) {
                dp[j] = Math.min(dp[j], 1 + dp[j-c]);
            }
        }
        return dp[amount] == amount + 2 ? -1 : dp[amount];
    }
}