161 - 322 零钱兑换
题目
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。
解答
感觉可以排序后从大到小减
大佬的套路太牛批了!!
总结一下:
动态规划算法是从暴力解优化而来的
优化就是用空间换时间。提高效率就是少做事
还可以自底向上,从一开始慢慢往后推延,就成了动态规划
动态规划
func min(a, b int) int {
if a < b {
return a
}
return b
}
func coinChange(coins []int, amount int) int {
dp := make([]int, amount+1)
for i := range dp {
dp[i] = amount + 1
}
dp[0] = 0
for i := 1; i <= amount; i++ {
for _, coin := range coins {
if coin <= i {
dp[i] = min(dp[i], dp[i-coin]+1)
}
}
}
if dp[amount] == amount+1 {
return -1
} else {
return dp[amount]
}
}
Runtime: 8 ms, faster than 90.91% of Go online submissions for Coin Change.
Memory Usage: 5.9 MB, less than 100.00% of Go online submissions for Coin Change.
class Solution:
def coinChange(self, coins: List[int], amount: int) -> int:
dp = [amount+1]*(amount+1)
dp[0] = 0
for i in range(amount+1):
for coin in coins:
if coin <= i:
dp[i] = min(dp[i], dp[i-coin]+1)
return -1 if dp[amount] == amount+1 else dp[amount]
Runtime: 1392 ms, faster than 53.04% of Python3 online submissions for Coin Change.
Memory Usage: 12.7 MB, less than 100.00% of Python3 online submissions for Coin Change.
var coinChange = function(coins, amount) {
const dp = new Array(amount + 1)
for (let i = 0; i < dp.length; i++) {
dp[i] = amount + 1
}
dp[0] = 0
for (let i = 0; i <= amount; i++) {
for (const coin of coins) {
if (coin <= i) {
dp[i] = Math.min(dp[i], dp[i - coin] + 1)
}
}
}
return dp[amount] == amount + 1 ? -1 : dp[amount]
};
Runtime: 108 ms, faster than 49.29% of JavaScript online submissions for Coin Change.
Memory Usage: 41.2 MB, less than 9.09% of JavaScript online submissions for Coin Change.
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