LeetCode 518. Coin Change 2
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原题链接在这里:https://leetcode.com/problems/coin-change-2/
题目:
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.
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
Note:
You can assume that
- 0 <= amount <= 5000
- 1 <= coin <= 5000
- the number of coins is less than 500
- the answer is guaranteed to fit into signed 32-bit integer
题解:
Let dp[i][j] denote with first i - 1 iron, to make up j, how many ways there are.
dp[i][j] = dp[i-1][j] + dp[i][j - coins[i - 1]].
dp[i - 1][j], without using coins[i - 1], the ways to make up j.
Since we are using coins[i - 1], we need to accumlate the ways from amount j - coins[i - 1].
Time Complexity: O(amount * n). n = coins.length.
Space: O(amount).
AC Java:
1 class Solution { 2 public int change(int amount, int[] coins) { 3 if(amount < 0 || coins == null){ 4 return 0; 5 } 6 7 int [] dp = new int[amount + 1]; 8 dp[0] = 1; 9 10 for(int coin : coins){ 11 for(int i = 1; i <= amount; i++){ 12 if(i - coin < 0){ 13 continue; 14 } 15 16 dp[i] += dp[i - coin]; 17 } 18 } 19 20 return dp[amount]; 21 } 22 }
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