输入：coins = [1, 2, 5], amount = 11输出：3 解释：11 = 5 + 5 + 1

输入：coins = [2], amount = 3输出：-1

输入：coins = [1], amount = 0输出：0

func coinChange(coins []int, amount int) int {    if len(coins) <= 0 {        return -1    }    if amount == 0 {        return 0    }
    sort.Slice(coins, func(i, j int) bool {        return coins[i] < coins[j]    })
    max := amount + 1    dp := make([]int, amount+1)    for i := 1; i <= amount; i++ {        dp[i] = max    }    for nowAmount := 1; nowAmount <= amount; nowAmount++ {        for i := 0; i < len(coins); i++ {            // 因为有了排序，所以一旦开始面值大于当前所求总和            // 后面的面值就不需要尝试了            if coins[i] > nowAmount {                break            }            if dp[nowAmount-coins[i]]+1 < dp[nowAmount] {                dp[nowAmount] = dp[nowAmount-coins[i]] + 1            }        }    }    if dp[amount] == max {        return -1    }    return dp[amount]}

func coinChange(coins []int, amount int) int {    if len(coins) <= 0 {        return -1    }    if amount == 0 {        return 0    }
    sort.Slice(coins, func(i, j int) bool {        return coins[i] > coins[j]    })
    res := amount + 1    var getNum func(coins []int, nowNum, remainRes int)    getNum = func(coins []int, nowNum, remainRes int) {        for i := 0; i < len(coins); i++ {            if coins[i] == remainRes && nowNum+1 < res {                res = nowNum + 1            } else if remainRes > coins[i] {                // 第二个剪枝，假设都用当前币种，所需数量依然大于现有结果，                // 那么就直接放弃，肯定结果数更大                if remainRes/coins[i] > res-nowNum {                    break                }                getNum(coins, nowNum+1, remainRes-coins[i])            }            // 第一个剪枝，对于超出结果的面值放弃，找更小面值硬币        }    }    getNum(coins, 0, amount)
    if res == amount+1 {        return -1    }    return res}