privateintres = Integer.MAX_VALUE; publicintcoinChange(int[] coins,intamount) { Arrays.sort(coins); coinChange(coins, amount, coins.length-1,0); returnres==Integer.MAX_VALUE ? -1: res; } // privatevoidcoinChange(int[] coins,intamount,intcur,intcount) { if(cur <0) return;// stop...
Input: coins = [1], amount = 2 Output: 2 Constraints: 1 <= coins.length <= 12 1 <= coins[i] <= 231 - 1 0 <= amount <= 104 classSolution {public://无限背包问题intcoinChange(vector<int>& coins,intamount) { vector<int> dp(amount+1,amount+1); dp[0] =0;for(inti=1;i<=...
建议和这一道题leetcode 518. Coin Change 2 动态规划DP 一起学习 这两到的DP的出发点是不一样的,本题是尽量凑够amout,然后求解最小的数量,518则是求解能够组成amount的所有的情况的计数,所以这两道题很类似,但是不一样 还建议和leetcode 279. Perfect Squares 类似背包问题 + 很简单的动态规划DP解决 一起...
1.) Minimum number of coins- Coin Change 1 on Leetcode 2.) Maximum number of ways- Coin Change 2 on Leetcode So, we have been given acoinsarray which consists of different denominations of the coins, and a totalamount. 1. Coin Change- 1 Here you need to find the minimum number of...
Breadcrumbs leetCode-4 /DP / CoinChange.pyTop File metadata and controls Code Blame 99 lines (76 loc) · 2.76 KB Raw """ You are given coins of different denominations and a total amount of money amount. Write a function to compute the fewest number of coins that you need to make ...
原题练习:LintCode 669.Coin Change 5最后总结 动态规划4步解题法 确定状态 研究最优策略的最后一步 转化为子问题 转移方程 根据子问题定义直接得到 初始条件和边界情况 细心,考虑周全 计算顺序 利用之前的计算结果 按照以上4步套路,基本上可以解决绝大多数类型的动态规划题。
visited[tem[0]+coin] = True if flag == False: # 当前层所有数字都比 amount 大,说明不可能组合成功 return -1 minl += 1 return -1 # 实际上不会执行这句话,因为这棵树是无穷尽的 print(Solution().coinChange([1,2,5], 11)) # 3...
def coinChange(self, coins, amount): """ :type coins: List[int] :type amount: int :rtype: int """ # 回溯法会超时 # 动态规划 dp = [-1]*(amount+1) dp[0]=0 for value in range(1,amount+1): for c in coins: if value-c>=0 and dp[value-c]!=-1: ...
class Solution: def coinChange(self, coins: List[int], amount: int) -> int: if amount < 0: return -1 if amount == 0: return 0 dp = [math.inf]*(amount+1) dp[0]=0 for coin in coins: for x in range(coin,amount+1): dp[x] = min(dp[x],1+dp[x-coin]) return dp[amou...
例子(leetcode-322): 解题思路: 对于每个金额,使用变量j遍历面值coins数组 publicintCoinChange(int[]coins,intamount){//定义一个数组,数组下标表示金额值(从0到amount),数据存放每种金额值所需的最少硬币个数(数组长度为amount+1,从0元开始)//例如amount=6,下标是: 0, 1, 2, 3, 4, 5, 6 , dp[6...