322. Coin Change

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. If that amount of money cannot be made up by any combination of the coins, return -1.

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

Example 1:

Input: coins = [1,2,5], amount = 11
Output: 3
Explanation: 11 = 5 + 5 + 1

Example 2:

Input: coins = [2], amount = 3
Output: -1

Example 3:

Input: coins = [1], amount = 0
Output: 0

Example 4:

Input: coins = [1], amount = 1
Output: 1

Example 5:

Input: coins = [1], amount = 2
Output: 2

Constraints:

1 <= coins.length <= 12
1 <= coins[i] <= 231 - 1
0 <= amount <= 104

递归解法

＃递归方法：＃
这个想法是非常经典的动态编程：想想我们要采取的最后一步。假设我们已经找到了总结，以量的最佳方法a，那么对于最后一步，我们可以选择任何硬币型这给了我们一个余数r，其中r = a-coins[i]所有i的。对于每个余数，请执行与之前完全相同的过程，直到余数为0或小于0（表示无效的解决方案）。有了这个想法，剩下的唯一细节就是存储总和所需的最小数量的硬币，r这样我们就不需要一遍又一遍地重新计算它。

public class Solution {
public int coinChange(int[] coins, int amount) {
    if(amount<1) return 0;
    return helper(coins, amount, new int[amount]);
}

private int helper(int[] coins, int rem, int[] count) { // rem: remaining coins after the last step; count[rem]: minimum number of coins to sum up to rem
    if(rem<0) return -1; // not valid
    if(rem==0) return 0; // completed
    if(count[rem-1] != 0) return count[rem-1]; // already computed, so reuse
    int min = Integer.MAX_VALUE;
    for(int coin : coins) {
        int res = helper(coins, rem-coin, count);
        if(res>=0 && res < min)
            min = 1+res;
    }
    count[rem-1] = (min==Integer.MAX_VALUE) ? -1 : min;
    return count[rem-1];
}
}

参考

https://leetcode.com/problems/coin-change/discuss/77368/Java-Both-iterative-and-recursive-solutions-with-explanations

动态规划解法

迭代方法：＃对于迭代解决方案，我们以自下而上的方式进行思考。假设我们已经计算出了所有的最小计数sum，那么最小计数将是sum+1多少？

public class Solution {
public int coinChange(int[] coins, int amount) {
    if(amount<1) return 0;
    int[] dp = new int[amount+1];
    int sum = 0;
    
	while(++sum<=amount) {
		int min = -1;
    	for(int coin : coins) {
    		if(sum >= coin && dp[sum-coin]!=-1) {
    			int temp = dp[sum-coin]+1;
    			min = min<0 ? temp : (temp < min ? temp : min);
    		}
    	}
    	dp[sum] = min;
	}
	return dp[amount];
}
}

标签：coin,min,int,coins,322,amount,rem,Coin,Change
来源： https://blog.csdn.net/zgpeace/article/details/116904673