322. Coin Change

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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.

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

Credits:
Special thanks to @jianchao.li.fighter for adding this problem and creating all test cases.

 

非常經典的dp問題

將amount作為dp的值，得到每個amount最少的硬幣數

init ： dp[0] = 0;

regular : dp[amount] = Math.min(dp[amount], dp[amount - coins[j]] + 1); 需設dp［i］＝ max;

o(n ^2);

//dp[amount] = Math.min(dp[amount], dp[amount - coins[j]] + 1);public 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];        dp[0] = 0;        for(int i = 1; i <= amount; i ++){            dp[i] = Integer.MAX_VALUE;            for(int j = 0 ; j < coins.length ; j++){                if(i >= coins[j] && dp[i - coins[j]] != Integer.MAX_VALUE)                    dp[i] = Math.min(dp[i], dp[i - coins[j]] + 1);            }        }        return dp[amount] == Integer.MAX_VALUE?-1:dp[amount];    }}

 

322. Coin Change