{LeetCode} 518. CoinChange2

问题：

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.

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

我的代码：

class Solution {public:    int change(int amount, vector<int>& coins) {        int count = 0;        if(amount==0) return 1;        else if(coins.empty()) return 0;        for(int i=0; i<coins.size();i++)        {            if(amount==coins.at(i))            {                count++;            }            else if(amount>coins.at(i))            {                vector<int> temp(coins);                vector<int>::iterator it;                for(it=temp.begin();it!=temp.end();)                {                    if(*it>coins.at(i))                    {                        it = temp.erase(it);                    }                    else{                        it++;                    }                }                count+=change(amount-coins.at(i),temp);            }        }        return count;    }};

结果：

Submission Result: Time Limit Exceeded

时间复杂度太高了，改用dynamic programming：

class Solution {public:    int change(int amount, vector<int>& coins) {        vector <vector<int> > dp(coins.size()+1, vector<int>(amount+1));        dp.at(0).at(0)=1;        for(int i=1;i<=coins.size();i++)        {            dp.at(i).at(0)=1;            for(int j=1;j<=amount;j++)            {                dp.at(i).at(j) = dp.at(i-1).at(j) + ((j-coins.at(i-1))>=0? dp.at(i).at(j-coins.at(i-1)):0);               }        }    return dp.at(coins.size()).at(amount);    }};