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
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)...
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: class Solution: def coinChange(self, coins: List[int], amount: int) -> int: n, m = len(coins), amount dp = [0 for i in range(m + 1)] for i in range(1,...
class Solution { public: int coinChange(vector<int>& coins, int amount) { const int N = coins.size(); vector<int> dp(amount + 1, INT_MAX); dp[0] = 0; for (int coin : coins) { for (int i = coin; i <= amount; ++i) { ...
LeetCode 322. Coin Change 题目 动态规划 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++)...
对已有的所有面值的硬币进行遍历,其实思路很简单:dp[i] += dp[i - coin],价值为i的解决方案应该加上价值为i - coin的解决方案。时间复杂度是O(L * A),空间复杂度是O(A); A = amount.代码如下:class Solution(object): def change(self, amount, coins): """ :type amount: int :type coins: ...
//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...
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...