sort((a,b)=>b-a) change(amount,0,0) return res === Infinity?-1:res // 内部调用函数类 function change(amount,count,cidx){ // base case if(amount === 0){ res = Math.min(count,res) return } if(coins.length === cidx) return let coin = coins[cidx] for(let i = Math....
leetcode:Coin Change You are given coins of different denominations and a total amount of moneyamount. Write a function to compute the fewest number of coins that you need to make up that amount. If that amount of money cannot be made up by any combination of the coins, return-1. Exampl...
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 of recursion coinChange inti = amo...
Coin Change 完全背包问题...完全可以推广到其它类型的背包问题,后面也就不再对进行状态转移之前的初始化进行讲解。 初始化情况一: LeetCode 322. Coin Change 完全背包问题,且要求正好装满(正好凑够amount钱 [leetcode]518. Coin Change 2 [leetcode]518. Coin Change 2 Analysis 还没工作就想退休—— [每天...
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func coinChange(coins []int, amount int) int { // dp[i] 表示凑出 i 所需的最少硬币数量, // 初始化为 amount + 1 ,表示当前还凑不出 dp := make([]int, amount + 1) for i := 0; i <= amount; i++ { dp[i] = amount + 1 } // 最开始只能确认不需要任何硬币就可以凑出 0 ...
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LeetCode: 322. Coin Change 题目描述 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 up that amount. If that amount of money cannot be made up by any combination of the coins...
Code Testcase Test Result Test Result 322. Coin Change Medium 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...
public int coinChange(int[] coins, int amount) { int[] state = new int[amount + 1]; for (int i = 1; i < state.length; i++) { int staMin = amount+1; for (int j = 0; j < coins.length; j++) { if (i - coins[j] >= 0 && state[i - coins[j]] != -1) { ...