public int coinChange(int[] coins, int amount) { if(coins==null||coins.length<=0 || amount<0) return -1; int[] dp=new int[amount+1]; //注意这里使用的是amount+1作为上限的 //不能用Integer.MAX_VALUE,因为会越界 Arrays.fill(dp, amount+1); dp[0]=0; for(int i=1;i<=amount;i...
Runtime:80ms, faster than22.84% of C++ online submissions for Coin Change. Memory Usage:14.5MB, less than27.45% of C++ online submissions for Coin Change. 三、优化措施 用dp写: classSolution{ public: //dp solution: dp[i] means the minimum num of coins used intcoinChange(vector<int>& co...
因为解法二用的是二维数组的dp,考虑将状态压缩成一维,于是就有第三种解法: 解法三: solve3 : dp3 var coinChange = function(coins,amount){ // 1,2,5 var dp = new Array(amount+1).fill(amount+1) dp[0] = 0 for(coin of coins){ for(var i = 0;i<=amount;i++){ if(i >= coin){ ...
dp初始状态:dp[0]=0,即当amount为0,以为coin的面值都是大于0的,所以不可能有任何任何一种硬币组合的面值为0 具体代码如下: classSolution:defcoinChange(self,coins:List[int],amount:int)->int:n,m=len(coins),amountdp=[0foriinrange(m+1)]foriinrange(1,m+1):min_value=1000forjinrange(n):# ...
我们设定的初始值dp=amount+1, 状态转移方程中,各个状态的dp都为初始值(未赋值的非法值时)min(dp+1,dp)则一定还为这个不可能取到的最大值。 代码参考: 1classSolution {2public:3//sub-problem:4//dp[s] = min(dp[s-coins[j]]+1)5//base:dp[0] = 06intcoinChange(vector<int>& coins,intamou...
if dp[-1] == float("inf"): return -1 else: return dp[-1] print(Solution().coinChange([4,3], 14)) # 4 方法2:广度优先搜索(BFS) 这道题和这样的题目很类似:一个人要从原点 0 走到 amount,每一次有不同走法,求走到 amount 所需要的最少步数。而之前写过这样的题目的做法:从M走到N最...
coin_change.py 源码 # 零钱兑换 from typing import List class Solution: def coinChange(self, coins: List[int], amount: int) -> int: dp = [0] + [amount + 1] * amount for i in range(1, amount + 1): for c in coins: if c <= i: dp[i] = min(dp[i], dp[i - c] + ...
class Solution { publicintchange(intamount,int[] coins){intn = coins.length;int[] dp =newint[amount +1]; dp[0] =1;// base casefor(inti =0; i < n; i++)for(intj =1; j <= amount; j++)if(j - coins[i] >=0) dp[j] = dp[j] + dp[j-coins[i]];returndp[amount]; ...
The Coin Changing problem The Coin Changing problem动态规划完美解决硬币找零问题
class Solution { public: int dp[10005]; int coinChange(vector<int>& coins, int amount) { memset(dp,-1,sizeof(dp)); dp[0] = 0; for(int i=1;i<=amount;i++) { for(int j=0;j<coins.size();j++) { if(i-coins[j]>=0 && dp[i-coins[j]]!=-1) ...