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| // @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]
}
|