LeetCode 322. Coin Change（java medium）

1.动态规划

将问题分解成子问题，自底向上的求解子问题的最优解，进而求出原问题的一个最优解（有可能有多个最优解，只求出其中的一个即可）。

2.题目

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.

3.分析

题目大意：给定硬币的不同面额和总钱数，用最少的硬币数凑齐总钱数，返回最少硬币数，不存在返回-1。

思路：利用动态规划的思想。利用数组dp[i]存储凑齐钱数i时所需硬币的最少数量，遍历coins计算dp[i]的值。coins数组是有序的

4.代码

public static int coinChange(int[] coins, int amount) {    int[] dp = new int[amount+1];    for (int i=1; i<=amount; i++) dp[i] = 0x7ffffffe;    for (int coin : coins) {        for (int i=coin; i<=amount; i++) {            dp[i] = Math.min(dp[i],dp[i-coin]+1);        }    }    return dp[amount]==0x7ffffffe ? -1 : dp[amount];}