在一个“零钱找零”类型的问题中,我试图将递归函数重构为迭代函数。给定一组硬币类型
coin_types
,函数 coinr
通过递归查找支付给定金额 sum
的最少硬币数量。# Any coin_type could be used more than once or it may not be used at all
sub coinr ($sum, @coin_types) { # As this is for learning basic programming
my $result = $sum; # No memoization (dynamic programming) is used
if $sum == @coin_types.any { return 1 }
else { for @coin_types.grep(* <= $sum) -> $coin_type {
my $current = 1 + coinr($sum - $coin_type, @coin_types);
$result = $current if $current < $result; } }
$result;
}
say &coinr(@*ARGS[0], split(' ', @*ARGS[1]));
# calling with
# 8 "1 4 5" gives 2 (4+4)
# 14 "1 3 5" gives 4 (1+3+5+5), etc.
这个函数最初是用Python编写的,我将其转换为了Raku。这是我的迭代版本,但它非常不完整:
# Iterative
sub coini ($sum, @coin_types) {
my $result = 1;
for @coin_types -> $coin_type {
for 1 ... $sum -> $i {
if $sum-$coin_type == @coin_types.any { $result += 1 } #?
else { # ???
}
}
}
}
有人可以帮我解决这个问题吗?
is copy
特性在函数内部使用参数。 - Lars Malmsteen