dp[i]表示凑成i所需的最少零钱个数
(1)先遍历物品，后遍历背包

func coinChange(coins []int, amount int) int {maxAmount := amount + 1dp := make([]int, amount+1)for i := 0; i <= amount; i++ {dp[i] = maxAmount}dp[0] = 0for _, coin := range coins {for i := coin; i <= amount; i++ {dp[i] = min(dp[i-coin]+1, dp[i])}}if dp[amount] != maxAmount {return dp[amount]} else {return -1}
}

(2)先遍历背包，后遍历物品

	for i := 1; i <= amount; i++ {for j := 0; j < len(coins); j++ {if i >= coins[j] {dp[i] = min(dp[i], dp[i-coins[j]]+1)}}}

dp[i]表示凑成平方和为i的最小元素个数
初始化：因为找最小个数，所以初始化为最大值，凑成平方和为0的最小元素个数为0
(1)先遍历物品，后遍历背包

func numSquares(n int) int {dp := make([]int, n+1)for i := 0; i <= n; i++ {dp[i] = math.MaxInt}dp[0] = 0for i := 1; i*i <= n; i++ {for j := i * i; j <= n; j++ {dp[j] = min(dp[j], dp[j-i*i]+1)}}return dp[n]
}

(2)先遍历背包，后遍历物品

	for i := 1; i <= n; i++ {for j := 1; j*j <= i; j++ {dp[i] = min(dp[i], dp[i-j*j]+1)}}