emu in blogjava

我的解法。
用的是笨笨的窮舉法~~
ps: Java的String 不能用[]，真不方便~~~
public class SecretSum
{
public int countPossible (String num1, String num2, String result)
{
int total = 0;
String s = num1 + num2 + result;
int i;
String uniqueChars = "";
char ch;
for (i = 0; i < s.length(); ++i)
{
ch = s.charAt(i);
if (uniqueChars.indexOf(ch) < 0)
uniqueChars += ch;
}

if (uniqueChars.length() > 10) // 10個以上不同的字母，出錯
return 0;

int maxTotal = 1; // 窮舉次數(shù)
for (i = 0; i < uniqueChars.length(); ++i)
maxTotal *= 10;

int numberMinValue = 1; // 每個數(shù)字最小值（用于限制最高位不能為0）
for (i = 1; i < num1.length(); ++i)
numberMinValue *= 10;

String temp; // 臨時變量
String s1, s2, s3; // 分別對應num1, num2, result，用于將字母替換成數(shù)字
int n1, n2, n3; // 替換后數(shù)字
boolean validNumber;
// 窮舉
for (int counter = maxTotal / 10; counter < maxTotal; ++counter)
{
temp = String.valueOf(counter);
if (temp.length() < uniqueChars.length())
continue;

validNumber = true;
for (i = 0; i < temp.length(); ++i)
{
if (temp.lastIndexOf(temp.charAt(i)) != i) // 計算每個數(shù)字是否只出現(xiàn)一次
validNumber = false;
}

if (validNumber)
{
s1 = num1;
s2 = num2;
s3 = result;
for (i = 0; i < temp.length(); ++i)
{
s1 = s1.replace(uniqueChars.charAt(i), temp.charAt(i));
s2 = s2.replace(uniqueChars.charAt(i), temp.charAt(i));
s3 = s3.replace(uniqueChars.charAt(i), temp.charAt(i));
}
n1 = Integer.valueOf(s1);
n2 = Integer.valueOf(s2);
n3 = Integer.valueOf(s3);
if (n1 >= numberMinValue && n2 >= numberMinValue && n1 + n2 == n3)
++total; // found 1
}
}
return total;
}

public static void main(String[] args)
{
SecretSum ss = new SecretSum();
System.out.println(ss.countPossible("AAA", "BBB", "CCC"));
System.out.println(ss.countPossible("ABB", "DEE", "TTT"));
System.out.println(ss.countPossible("ABC", "ABA", "ACC"));
System.out.println(ss.countPossible("AAA", "CDD", "BAA"));
System.out.println(ss.countPossible("TEF", "FET", "AAA"));
System.out.println(ss.countPossible("ABC", "ABC", "BCE"));
System.out.println(ss.countPossible("AAA", "AAA", "BBB"));
}
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if (temp.lastIndexOf(temp.charAt(i)) != i) validNumber = false;
這句忘了加個 break;
不影響結(jié)果，不過效率高點點..  回復  更多評論
  

很不錯的算法,樓主厲害啊  回復  更多評論
  

如果有更多朋友寫答案上來就更好了。  回復  更多評論
  

@Tendy
  回復  更多評論
  

"AAA"
"CDD"
"BAA"

Returns: 32

我編的算法，做出來怎 么是31？指點下…看看有什么疏忽…
111+300=411
111+400=511
111+500=611
111+600=711
111+700=811
111+800=911
222+100=322
222+300=522
222+400=622
222+500=722
222+600=822
222+700=922
333+100=433
333+200=533
333+400=733
333+500=833
333+600=933
444+100=544
444+200=644
444+300=744
444+500=944
555+100=655
555+200=755
555+300=855
555+400=955
666+100=766
666+200=866
666+300=966
777+100=877
777+200=977
888+100=988
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