Middle-题目94:322. Coin Change
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题目原文:
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.
题目大意:
假设你有若干面额的钱币,面额由数组coins给出,再给出一个总的钱币数amount。计算至少需要几枚钱币才能凑成amount?如果能,返回最小钱数,如果不能,返回-1.
例如:coins = [1, 2, 5], amount = 11
return 3 (11 = 5 + 5 + 1)
题目分析:
使用一个
即已知钱数i所需要的硬币数,可以推出再加一枚硬币可以表示的钱数.最终dp[amount]即为所求。(这时候看出来为什么写了个-1吗?)
源码:(language:java)
public class Solution { public int coinChange(int[] coins, int amount) { int dp[] = new int[amount + 1]; for (int i = 1; i <= amount; i++) dp[i] = Integer.MAX_VALUE-1; for (int i = 0; i <= amount; i++) { for (int j = 0; j < coins.length; j++) { if (i + coins[j] <= amount) dp[i + coins[j]] = Math.min(dp[i + coins[j]], dp[i] + 1); } } return dp[amount] == Integer.MAX_VALUE-1 ? -1 : dp[amount]; }}
成绩:
23ms,beats 83.94%,众数25ms,9.55%
Cmershen的碎碎念:
读完代码之后应该能解释这个-1了。如果还没理解那么揭晓答案:计算机中用补码储存数字,所以如果不写-1,那么在第9行判断dp[i]+1的时候会造成溢出,导致变成MIN_VALUE,那么dp[i+coins[j]]不管是多少都会被冲掉。
此外这题好像可以用母函数在O(n)复杂度内做,我组合数学学得不好,也没查到具体的ac代码。(这题又是那个fighter.io出的,貌似他出的题总离不开数学)
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