(Java) LeetCode 322. Coin Change —— 零钱兑换

标签：tco   mon   不能   ota   ann   inf   inpu   may   通用   

You are given coins of different denominations and a total amount of money amount. Write a function to compute 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.

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

Note:
You may assume that you have an infinite number of each kind of coin.

动态规划典型问题，想一想递推关系是什么。假设dp[i]代表i这么多钱所需要的硬币数量，那么对于每一个不同面值的硬币coins[j]来说，dp[i] = dp[i - coins[j]] + 1。只要在遍历coins面值的时候选择需要数量最小的就好，即min(dp[i], dp[i - coins[j]] + 1)。通用情况想完了就要想初始情况。dp[0]显然是0，不需要硬币组成0元。而既然在递推中要用到min，那么把所有值初始为amount + 1的话即可，因为硬币都是正数，不可能会存在需要比amount还多硬币的情况。如果最后得到的结果比amount大（其实题里并没有说amount不会是最大正数，否则这样取值就会overflow产生错误，test case里面并没有这种情况），证明没有办法组成面值，需要返回-1。需要注意的是因为i-coins[j]作为数组下标出现，显然conis[j]不能比i大。

Java

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

(Java) LeetCode 322. Coin Change —— 零钱兑换

标签：tco   mon   不能   ota   ann   inf   inpu   may   通用   

原文地址：https://www.cnblogs.com/tengdai/p/9249164.html