LeetCode刷题之322 Coin Change

1. Description

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.

2. Test Cases

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

3. Solution

3.1 带备忘录递归

class Solution {
    
    public int coinChange(int[] coins, int amount) {
    
        Integer[] dp = new Integer[amount +1];
        dp[0] = 0;
        
        find(coins, amount, dp);
        
        return dp[amount];
    }
    
    private int find(int[] coins, int amount, Integer[] dp) {
    
        if (amount == 0) return 0;
        
        if(dp[amount] != null )
            return dp[amount];
        
        int min =Integer.MAX_VALUE;
        
        for (int c : coins) {
    
            if (amount -c >= 0) {
    
                int r = find(coins, amount-c, dp);
                if (r != -1)
                    min = Math.min(min, r + 1);
            }
        }
        
        return dp[amount] = (min == Integer.MAX_VALUE ? -1 : min);
    }
    
}

3.2 iterative

class Solution {
    
    public int coinChange(int[] coins, int amount) {
    
        Integer[] dp = new Integer[amount +1];
        dp[0] = 0;
        
        for (int i = 1; i <= amount; i++) {
    
            int min = Integer.MAX_VALUE;
            
            for (int c : coins) {
    
                if (i - c >= 0 ) {
    
                    int r = dp[i - c];
                    if (r != -1)
                        min = Math.min(min, r + 1);
                }
            }
            
            dp[i] = min == Integer.MAX_VALUE ? -1 : min;
        }
        
        return dp[amount];
    }
}

4. 思考

似乎动态规划的本质就是穷举，只是如何穷举更少的次数，更快的得到结果。