python---通过递归和动态规划策略解决找零钱问题

也是常见套路。

# coding = utf-8

def rec_mc(coin_value_list, change, know_results):
    min_coins = change
    if change in coin_value_list:
        know_results[change] = 1
        return 1
    elif know_results[change] > 0:
        return know_results[change]
    else:
        for i in [c for c in coin_value_list if c <= change]:
            num_coins = 1 + rec_mc(coin_value_list, change-i, know_results)
            if num_coins < min_coins:
                min_coins = num_coins
                know_results[change] = min_coins
    return min_coins

print("===========递归实现========================")
print(rec_mc([1, 5, 10, 25], 63, [0]*64))

def dp_make_change(coin_value_list, change, min_coins, coins_used):
    for cents in range(change+1):
        coin_count = cents
        new_coin = 1
        for j in [c for c in coin_value_list if c <= cents]:
            if min_coins[cents-j] + 1 < coin_count:
                coin_count = min_coins[cents-j]+1
                new_coin = j
        min_coins[cents] = coin_count
        coins_used[cents] = new_coin
    return min_coins[change]

def print_coins(coins_used, change):
    coin = change
    while coin > 0:
        this_coin = coins_used[coin]
        print(this_coin)
        coin = coin - this_coin

a_mnt = 63
c_list = [1, 5, 10, 21, 25]
c_used = [0] * (a_mnt+1)
c_count = [0] * (a_mnt+1)
print("===========动态规划实现========================")
print('Making change for ', a_mnt, 'requires')
print(dp_make_change(c_list, a_mnt, c_count, c_used), 'coins')
print("They are: ")
print_coins(c_used, a_mnt)
print("The used list is as follows: ")
print(c_used)

　　

输出：

D:\cheng\test\Scripts\python.exe tests.py
===========递归实现========================
6
===========动态规划实现========================
Making change for  63 requires
3 coins
They are:
21
21
21
The used list is as follows:
[1, 1, 1, 1, 1, 5, 1, 1, 1, 1, 10, 1, 1, 1, 1, 5, 1, 1, 1, 1, 10, 21, 1, 1, 1, 25, 1, 1, 1, 1, 5, 10, 1, 1, 1, 10, 1, 1, 1, 1, 5, 10, 21, 1, 1, 10, 21, 1, 1, 1, 25, 1, 10, 1, 1, 5, 10, 1, 1, 1, 10, 1, 10, 21]

Process finished with exit code 0