每日一题 day 58(DP topic)

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#动态规划 #leetcode #算法

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本文探讨了518.Coin Change 2问题的解决方法。该问题要求计算出组成特定金额的不同硬币组合数量。文章首先介绍了递归方法，并分析了其不足之处；随后提出了一种使用记忆化的动态规划方法，有效地解决了问题。

文章目录

problem
wrong approach
approach `memorize dp`

problem

518. Coin Change 2
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

wrong approach

class Solution {
public:
    vector<vector<int>> dp;
    int rec(int amount, int depth, vector<int>& coins){
        if(amount < 0)return 0;
        if(amount == 0){
            // if(dp[depth] == 0)
            //     return dp[depth] = 1;
            // else return 0;
            return 1;
        }
        //if(dp[depth + amount/coins[0]] == 1) return 0;
        
        int res = 0;
        for(auto coin : coins){// iterate through all kinds of coin
            res += rec(amount - coin, depth+1, coins);// tmp = subtree's return
        }
        return res;
    }
    int change(int amount, vector<int>& coins) {
        if(coins.size() == 0) return 0;
        sort(coins.begin(), coins.end());
        dp = vector<int>((amount/coins[0]) + 1, 0);
        return rec(amount, 0, coins);
    }
};

approach `memorize dp`

class Solution {
public:
    vector<vector<int>> dp;
    int solve(vector<int>& coins,int m,int n){
        if(m==-1 && n>0) 
            return 0;
        if(n==0) 
            return 1;
        if(n<0) 
            return 0;
        if(dp[m][n]!=-1) 
            return dp[m][n];
        return dp[m][n]=solve(coins,m,n-coins[m])+solve(coins,m-1,n);
    }

    int change(int amount, vector<int>& coins) {
        dp = vector<vector<int>>(coins.size(), vector<int>(amount+1, -1));
        return solve(coins,coins.size()-1,amount);
    }
};