【LeetCode】322. Coin Change 零钱兑换（Medium）（JAVA）
题目地址: https://leetcode.com/problems/coin-change/
题目描述：
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.
You may assume that you have an infinite number of each kind of coin.
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
Example 3:
Input: coins = [1], amount = 0
Output: 0
Example 4:
Input: coins = [1], amount = 1
Output: 1
Example 5:
Input: coins = [1], amount = 2
Output: 2
Constraints:
- 1 <= coins.length <= 12
- 1 <= coins[i] <= 2^31 - 1
- 0 <= amount <= 10^4
题目大意
给定不同面额的硬币 coins 和一个总金额 amount。编写一个函数来计算可以凑成总金额所需的最少的硬币个数。如果没有任何一种硬币组合能组成总金额，返回 -1。
你可以认为每种硬币的数量是无限的。
解题方法
- 开始采用暴力搜索 BFS（宽度优先搜索），直接超时
- 用动态规划，最重要的是找出状态转移函数 dp 函数：dp[i] = Math.min(dp[i], dp[i - coins[j]] + 1);
- note: 因为可能不存在，所以要额外定义一个不存在的状态，比如 dp[i] = -1
class Solution {
public int coinChange(int[] coins, int amount) {
int[] dp = new int[amount + 1];
Arrays.sort(coins);
for (int i = 1; i <= amount; i++) {
dp[i] = -1;
for (int j = 0; j < coins.length; j++) {
if (i - coins[j] < 0 || dp[i - coins[j]] < 0) continue;
if (dp[i] == -1 || dp[i] > (dp[i - coins[j]] + 1)) {
dp[i] = dp[i - coins[j]] + 1;
}
}
}
return dp[amount];
}
}
执行耗时:18 ms,击败了37.63% 的Java用户
内存消耗:38.1 MB,击败了54.99% 的Java用户