1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59

// @Title: 零钱兑换 (Coin Change)
// @Author: 15816537946@163.com
// @Date: 2019-09-22 18:05:21
// @Runtime: 12 ms
// @Memory: 5.8 MB
/*
 * @lc app=leetcode.cn id=322 lang=golang
 *
 * [322] 零钱兑换
 *
 * https://leetcode-cn.com/problems/coin-change/description/
 *
 * algorithms
 * Medium (34.89%)
 * Likes:    210
 * Dislikes: 0
 * Total Accepted:    19.6K
 * Total Submissions: 56.3K
 * Testcase Example:  '[1,2,5]\n11'
 *
 * 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。
 * 
 * 示例 1:
 * 
 * 输入: coins = [1, 2, 5], amount = 11
 * 输出: 3 
 * 解释: 11 = 5 + 5 + 1
 * 
 * 示例 2:
 * 
 * 输入: coins = [2], amount = 3
 * 输出: -1
 * 
 * 说明:
 * 你可以认为每种硬币的数量是无限的。
 * 
 */
func coinChange(coins []int, amount int) int {
	dp := make([]int, amount+1)
	dp[0] = 0

	for i := 1;i<=amount;i++ {
		dp[i] = amount+1
		for _, c := range  coins {
			if c <= i && dp[i] > dp[i-c] +1 {
				dp[i] = dp[i-c] +1
			}
		}
	}

	if dp[amount] > amount {
		return -1
	}

	return dp[amount]
}