LeetCode 322. Coin Change

原題連接在這裏：https://leetcode.com/problems/coin-change/

題目：

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:
coins = [1, 2, 5], amount = 11
return 3 (11 = 5 + 5 + 1)

Example 2:
coins = [2], amount = 3
return -1.

題解：

要求知足amount時用的最少coins數目.

Let dp[i] denotes amount == i 時用的最少coins數目. 用一維array儲存.

遞推時，狀態轉移dp[i] = Math.min(dp[i], dp[i-coins[j]]+1).  dp[i-coins[j]]+1表示用最少個數coin表示i-coins[j].

初始化dp[0] = 0. amount爲0時不須要硬幣. dp其餘值都是Integer的最大值, 以後會每次更新取最小值.

Time Complexity: O(amount * coins.length).

Space: O(amount).

AC Java:

class Solution {
    public int coinChange(int[] coins, int amount) {
        if(amount == 0){
            return 0;
        }
        
        if(coins == null || coins.length == 0){
            return -1;
        }
        
        int [] dp = new int[amount+1];
        dp[0] = 0;
        for(int i = 1; i<=amount; i++){
            dp[i] = Integer.MAX_VALUE;
            for(int coin : coins){
                if(i-coin<0 || dp[i-coin]==Integer.MAX_VALUE){
                    continue;
                }else{
                    dp[i] = Math.min(dp[i], dp[i-coin]+1);
                }
            }
        }
        
        return dp[amount] == Integer.MAX_VALUE ? -1 : dp[amount];
    }
}

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