当前位置:   article > 正文

518. Coin Change II

518. Coin Change II

You are given an integer array coins representing coins of different denominations and an integer amount representing a total amount of money.

Return the number of combinations that make up that amount. If that amount of money cannot be made up by any combination of the coins, return 0.

You may assume that you have an infinite number of each kind of coin.

The answer is guaranteed to fit into a signed 32-bit integer.

Example 1:

Input: amount = 5, coins = [1,2,5]
Output: 4
Explanation: there are four ways to make up the amount:
5=5
5=2+2+1
5=2+1+1+1
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
  1. class Solution {
  2. public:
  3. int change(int amount, vector<int>& coins) {
  4. vector<int> dp(amount + 1);
  5. dp[0] = 1;
  6. for (int& coin : coins) {
  7. for (int i = coin; i <= amount; i++) {
  8. dp[i] += dp[i - coin];
  9. }
  10. }
  11. return dp[amount];
  12. }
  13. };

声明:本文内容由网友自发贡献,不代表【wpsshop博客】立场,版权归原作者所有,本站不承担相应法律责任。如您发现有侵权的内容,请联系我们。转载请注明出处:https://www.wpsshop.cn/w/盐析白兔/article/detail/295103
推荐阅读
  

闽ICP备14008679号