同时还有一个细节:在设置f的长度的时候要设置成amount + 1,因为这里是1 indexed(index对应的即选择的amount,而不是amount - 1)。 解法1: 1#definevi vector<int>2classSolution {3public:4intcoinChange(vi& coins,intamount) {5vi f(amount, amount +1);6intn =coins.size();7f[0] =0;8for(inti...
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 ...
public int coinChange(int[] coins, int amount) { if (amount < 0) return 0; return coinChangeCore(coins, amount, new int[amount]); } private int coinChangeCore(int[] coins, int amount, int[] count) { if (amount < 0) return -1; if (amount == 0) return 0; //剪枝 if (coun...
# Assume at first that all 1's are used.dp[i]=i 状态转移方程 如果当前的钱数i,比coins中的一些硬币大的话,那么就看一下mincoins[i]和mincoins[i-c]+1哪一个更大,取其中的最小值。 # Check if any coin leads to a better solution to our current best.forcincoins:ifi>=c:dp[i]=min(dp...
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322. Coin ChangeMedium Topics Companies You are given an integer array coins representing coins of different denominations and an integer amount representing a total amount of money. Return the fewest number of coins that you need to make up that amount. If that amount of money cannot be made...
本⽂聊的是 LeetCode 第 518 题 Coin Change 2,题⽬如下: Leetcode算法题第518题 PS:⾄于 Coin Change 1,在我们前⽂ 动态规划套路详解 写过。 我们可以把这个问题转化为背包问题的描述形式: 有⼀个背包,最⼤容量为 amount ,有⼀系列物品 coins ,每个物品的重量为 coins[i] ,每个物品的数量...
1 2 3 4 5 6 classSolution{ public: intcoinChange(vector<int>&coins,intamount) { } }; 已存储 行1,列 1 运行和提交代码需要登录 coins = [1,2,5] amount = 11 9 1 2 3 4 5 6 › [1,2,5] 11 [2] 3 [1] 0 Source
func coinChange(coins []int, amount int) int { //dp[i]: amount 的钱需要的最少硬币数 (-1 表示不可达) dp := []int {0} for i := 1; i <= amount; i++ { dp = append(dp, -1) for j := 0; j < len(coins); j++ { ...
ChangeDFS(vector<int> &coins,intamount, vector<int> &dp) {if(amount <0)return-1;if(dp[amount] != INT_MAX)returndp[amount];for(inti =0; i < coins.size(); ++i) {inttmp = coinChangeDFS(coins, amount -coins[i], dp);if(tmp >=0) dp[amount] = min(dp[amount], tmp +1);...