迭代法塔的问题，如果中国

看递归解决方案。使用Perl语言完成了不到一分钟。

sub hanno_recursive {
my ($from, $to, $reserve, $n) =  @_;
if (1 == $n) {
print "move $n from $from to $to\n";
return;
}

hanno_recursive($from, $reserve, $to, $n -1);
print "move $n from $from to $to\n";
hanno_recursive($reserve, $to, $from, $n -1);
}

极其简洁优美。充分体现了递归的优雅。

接下来。考虑迭代解法。考虑将问题分解为树结构。非常显然， 将A   B  C看成一个圈， 则左右子树具有某种对称。即顺时针或逆时针旋转。

这样。我们全然能够通过左树求得右树， 问题变成为了线性递归， 这个非常easy转换为迭代。

原理知道了， 写这个代码，还是比較费劲， 花了1个小时才调好。

sub hanno_iterate {
my ($from, $to, $reserve, $n) =  @_;
my @left = ();
my @right = ();
#move to leaf node
my $count = $n;
while ($count > 1) {
my $tmp = $to;
$to = $reserve;
$reserve = $tmp;
$count--;
}

for (my $index = 1; $index <= $n; $index++) {
my $new = "move $index from $from to $to\n";

push @left, $new;
while ($new = shift @right) {
push @left, $new;
}
last if ($index == $n);

if (($index % 2) == ($n % 2)){
# anti-clock $from -> $to, $reserve->$from, $to -> $reserve
foreach my $opt(@left) {
my $left_value = "$opt";
$left_value =~ tr/ABC/CAB/;
push @right, $left_value;
}
} else {
# clock $from -> $reserve, $reserve->$to, $to -> $from
foreach my $opt(@left) {
my $left_value = "$opt";
$left_value =~ tr/ABC/BCA/;
push @right, $left_value;
}
}
my $tmp = $to;
$to = $reserve;
$reserve = $tmp;
}

foreach my $opt(@left) {
print $opt;
}


}

版权声明：本文博客原创文章，博客，未经同意，不得转载。

转载于:https://www.cnblogs.com/yxwkf/p/4627868.html