最少硬币数

#include <iostream>

#include <set>

#include <thread>

#include <mutex>

#include <deque>

#include <algorithm>

using namespace std;

/*给定不同面额的硬币 coins 和一个总金额 amount。编写一个函数来计算可以凑成总金额所需的最少的硬币个数。

如果没有任何一种硬币组合能组成总金额，返回-1。

假设每一种面额的硬币有无限个。



输入: coins = [1, 3, 7], amount = 9

输出: 3

解释:

3 -> 9 = 3×3 = 1+1+7

9 -> 9 = 1×9*/



int bt(vector<int>& coins, int val, int l, int r) {

if (val == 0) {

return 0;

}



while (r >= l) {

if (coins[r] <= val) {

break;

}

r--;

}



if (r < l) {

return -1;

}



cout << "tmp_val :" << coins[r] << endl;

auto res = bt(coins, val - coins[r], l, r);

if (res == -1) {

return -1;

}



return 1 + res;



}

int calMinNum(vector<int>& coins, int amount) {

sort(coins.begin(), coins.end());



int l = 0;

int r = coins.size() - 1;



return bt(coins, amount, l, r);

}




int main() {

vector<int> coins = {1, 2, 7};

int amount = 9;



auto res = calMinNum(coins, amount);

cout << "res:" << res << endl;

return 0;

}