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_322.java
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_322.java
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package com.fishercoder.solutions;
import java.util.Arrays;
/**
* 322. Coin Change
*
* 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, return -1.
Example 1:
coins = [1, 2, 5], amount = 11
return 3 (11 = 5 + 5 + 1)
Example 2:
coins = [2], amount = 3
return -1.
Note:
You may assume that you have an infinite number of each kind of coin.
*/
public class _322 {
public static class Solution1 {
public int coinChange(int[] coins, int amount) {
if (amount < 1) {
return 0;
}
int[] count = new int[amount];
int result = helper(coins, amount, count);
return result;
}
//remaining means the remaining coins after the last step;
//count[remaining] means the minimum number of coins to sum up to remaining
private int helper(int[] coins, int remaining, int[] count) {
if (remaining < 0) {
return -1;//not valid case, thus, per problem description, we should return -1
}
if (remaining == 0) {
return 0;//completed, this is also a base case for this recursive function
}
if (count[remaining - 1] != 0) {
return count[remaining - 1];//already computed, so reuse it.
}
int min = Integer.MAX_VALUE;
for (int coin : coins) {
int res = helper(coins, remaining - coin, count);
if (res >= 0 && res < min) {
min = 1 + res;
}
}
return count[remaining - 1] = (min == Integer.MAX_VALUE) ? -1 : min;
}
}
public static class Solution2 {
//dp solution
public int coinChange(int[] coins, int amount) {
int max = amount + 1;
int[] dp = new int[max];
Arrays.fill(dp, max);// initial the dp array with amount + 1 which is not valid case.
dp[0] = 0;//initial amount 0 = 0;
for (int i = 1; i <= amount; i++) {
for (int j = 0; j < coins.length; j++) {
if (coins[j] <= i) {
//the dp[coins[j]] will ba a valid case, then if dp[i - coins[j]] is valid
dp[i] = Math.min(dp[i], dp[i - coins[j]] + 1);
// then we update dp[i], otherwise dp[i] = max;
}
}
}
return dp[amount] > amount ? -1 : dp[amount];
}
}
}