图片来自于https://leetcode-cn.com/problems/coin-change/solution/wan-quan-bei-bao-wen-ti-shou-hua-dp-table-by-shixu/ 值得一提的是, 以原问题 amount = 11, coins = [2,5,1] 为例子。 i=1代表的是只可以放 {coin = 2,count =1}下,可以放n个小于11的情况 ...
This is a classic problem and greedy might be the first thing comes into mind. For example, sort the coins on denomination and always use the largest amount coin. However, the optimal solution might not include the largest denomination. for example , [1, 3, 4] and target value 6. The ...
https://leetcode-cn.com/problems/coin-change 解题思路 动态规划,自底向上,太简单,不解释。 C++代码 classSolution{public:intcoinChange(vector<int>& coins,intamount){vector<int>dp(amount +1, amount +1); dp[0] =0;for(inti =1; i <= amount; i++) {for(autov : coins) {if(i >= v)...
吐槽一下leetcode怎么有这么多诡异的testcase。 classSolution:defcoinChange(self,coins:List[int],amount:int)->int:ifamount==0:return0dp=[amount+1]*(amount+1)dp[0]=0foriinrange(1,amount+1):forcincoins:ifi>=c:dp[i]=min(dp[i],dp[i-c]+1)returndp[-1]ifdp[-1]!=amount+1else-1 ...
public class Solution { public int coinChange(int[] coins, int amount) { if(coins.length==0) return -1; if(amount==0) return 0; int[] dp=new int[amount+1]; //start from 1, intead of 0 for(int i=1;i<=amount;i++){ int minNumber=Integer.MAX_VALUE; for(int j=0;j<coins...
classSolution{public:intcoinChange(vector<int>&coins,int amount){int size=coins.size();//获取硬币个数//因为有个初始状态0,即什么硬币都不考虑,因此行数加一行vector<vector<int>>dp(size+1,vector<int>(amount+1,0));// 初始化(没有任何硬币的情况):只有 f[0][0] = 0;其余情况均为无效值。
class Solution: def coinChange(self, coins: List[int], amount: int) -> int: dp=[amount+1]*(amount+1) dp[0]=0 for i in range(1,amount+1): for c in coins: if i>=c: dp[i]=min(dp[i],dp[i-c]+1) if dp[amount]==amount+1: ...
class Solution {public: int coinChange(vector<int>& coins, int amount) { auto dp = vector<int>(amount + 1, INT_MAX); dp[0] = 0; for (auto a = 1; a <= amount; ++a) { for (const auto & coin : coins) { if (a >= coin && dp[a - coin] < INT_MAX...
class Solution: def change(self, amount: int, coins: List[int]) -> int: dp = [0] * (amount + 1) dp[0] = 1 for j in range(len(coins)): for i in range(1, amount + 1): if i >= coins[j]: dp[i] += dp[i - coins[j]] return dp[-1]复杂度...
//RecursionclassSolution {public:intcoinChange(vector<int>& coins,intamount) { unordered_map<int,int>dp; dp[0] =0;returncoinChangeDFS(coins, amount, dp); }intcoinChangeDFS(vector<int> &coins,intamount, unordered_map<int,int> &dp) {if(amount <0)return-1;if(dp.find(amount) != dp...