C++int change(int amount, vector<int>& coins) { vector<int> dp(amount + 1); dp[0] = 1; for (auto& coin : coins) { for (int i = coin; i <= amount; i++) { dp[i] += dp[i - coin]; } } return dp[amount]; } 分类: LeetCode 标签: 动态规划 好文要顶 关注我 收藏...
5=1+1+1+1+1 Example 2: Input: amount = 3, coins = [2] Output: 0 Explanation: the amount of 3 cannot be made up just with coins of 2. Example 3: Input: amount = 10, coins = [10] Output: 1 Note: You can assume that 0 <= amount <= 5000 1 <= coin <= 5000 the numbe...
1classSolution {2publicintchange(intamount,int[] coins) {3if(amount < 0 || coins ==null){4return0;5}67int[] dp =newint[amount + 1];8dp[0] = 1;910for(intcoin : coins){11for(inti = 1; i <= amount; i++){12if(i - coin < 0){13continue;14}1516dp[i] += dp[i -co...
Coin Change 2 https://leetcode.com/problems/coin-change-2/description/ 完全背包吧,然后两重循环的次序需要注意。 最外层一定是coins 如果外层是amount,那么比如amount=3会在coin1和coin2分别+=dp[1], +=dp[2] 这两种其实都是2+1的情况 导致重复......
题目链接: https://leetcode.com/problems/coin-change-2/description/ 题解: 完全背包 代码:...leetcode 518. Coin Change 2 一、题意 给一个整数数组 coins 表示不同面额的硬币,另给一个整数 amount 表示总金额。 请计算并返回可以凑成总金额的硬币组合数。如果任何硬币组合都无法凑出总金额,返回 0 。
所有的动态规划问题都可以转化为暴力递归的过程,此类类型的暴力模型是: for(状态1 in 所有取值){ for(状态2 in 所有取值){ dp[状态1][状态2] = 计算(选择1, 选择2) } } 下面解法使用了两种数组:一维数组、二维数组,一维数组更好理解,二维数组空间复杂度稍好,时间复杂度都是O(n*amount)...
// LeetCode #272 medium // 518. Coin Change 2 // Runtime: 72 ms, faster than 12.21% of C++ online submissions for Coin Change 2. // Memory Usage: 22.2 MB, less than 14.29% of C++ online submissions …
【DP】518. Coin Change 2 问题描述: You are given coins of different denominations and a total amount of money. Write a function to compute the number of combinations that make up that amount. You may assume that you have infinite number of each kind of coin....
PURPOSE: To obtain high discharging performance capable of discharging quickly and surely change to a receiving tray by providing a temporary stop control means to stop temporarily the carrying operation of a change conveyor in the course of carrying.IKEZAWA TOSHIAKI池澤 利明...
int main() { int n, d[251] = {0}; int c1, c5, c10, c25, c50; for (n = 0; n<251;n++) for(c50=0;c50*50<= n; c50++) for (c25 = 0; c50 * 50 + c25 * 25<=n;c25++) for(c10=0;c50*50+c25*25+c10*10<= n; c10++) ...