Question

我在iOS工作，我遇到了一个问题。我有6个字母，我想找到这6个字母可以生成的所有字符串。字符串的序列和长度在这里无关紧要，字符串的含义也无关紧要，因为我们可以用字典检查字符串，以确定它是否是有效字。

那么有什么算法可以找到它吗？

例如：

输入6个字母：N T L P A E

预期输出将是：

plan
net
planet
lan
tea
lap
..
..

这里输出中的单词只是有效单词，但输出应该包含所有可能的单词，甚至是无效单词。

Answer 1

这应该可以解决这个问题：

+ (void) logPermutations:(NSArray*)objects
{
    if (objects == nil || [objects count] == 0) {
        return;
    }
    NSMutableArray* copy = [objects mutableCopy];
    [self logPermutations:copy logedSoFar:@""];
}

+ (void) logPermutations:(NSMutableArray*)objects
              logedSoFar:(NSString*)log
{
    if (objects == nil || [objects count] == 0) {
        return;
    }
    NSUInteger count = [objects count];
    for (NSUInteger i = 0; i < count; ++i) {
        id removed = [objects objectAtIndex:i];
        [objects removeObjectAtIndex:i];
        NSString* newlog = [NSString stringWithFormat:@"%@%@",log,[removed description]];
        NSLog(@"%@",newlog);
        [self logPermutations:objects logedSoFar:newlog];
        [objects insertObject:removed atIndex:i];
    }
}

Answer 2

-(NSMutableArray*)genValidWords:(NSString*)tiles
{
    /*===============EMPTY PREVIOUS ARRAY=================*/
    FINAL_VALID_WORDS=[[NSMutableArray alloc] init];
    posWords=[[NSMutableArray alloc] init];
    genWords=[[NSArray alloc] init];
    /*============EMPTY PREVIOUS ARRAY FINISH=============*/

    /*===============POSSIABLE NO OF WORDS FROM GIVEN LETTERS=================*/

    int length=[tiles length];
    int total=0;
    for(long i=2;i<[tiles length]+1;i++)
        total=total+([self factorialX:length]/[self factorialX:length-i]);
    NSLog(@"POSSIABLE NO OF WORDS FROM LETTERS \"%@\" IS %d",tiles,total);

    /*===============POSSIABLE NO OF WORDS FROM GIVEN LETTERS FINISH=================*/



    /*===============GET ARRAY OF ALL POSSIABLE WORDS FROM GIVEN LETTERS=============*/
    for(int i=2;i<[tiles length]+1;i++)
    {
        genWords=getPermutations(tiles, i);
        [posWords addObjectsFromArray:genWords];
     }
     NSLog(@"%@",posWords);
    return posWords;
}
static NSMutableArray *results;
void doPermute(NSMutableArray *input, NSMutableArray *output, NSMutableArray *used, int size, int level)
{
    if (size == level)
    {
        NSString *word = [output componentsJoinedByString:@""];
        //if(check(word))
        [results addObject:word];
        return;
    }
    level++;
    for (int i = 0; i < input.count; i++)
    {
        if ([used[i] boolValue])
        {
            continue;
        }
        used[i] = [NSNumber numberWithBool:YES];
        [output addObject:input[i]];
        doPermute(input, output, used, size, level);
        used[i] = [NSNumber numberWithBool:NO];
        [output removeLastObject];
    }
}
NSArray *getPermutations(NSString *input, int size)
{
    results = [[NSMutableArray alloc] init];

    NSMutableArray *chars = [[NSMutableArray alloc] init];


    for (int i = 0; i < [input length]; i++)
    {
        NSString *ichar  = [NSString stringWithFormat:@"%c", [input characterAtIndex:i]];
        [chars addObject:ichar];
    }

    NSMutableArray *output = [[NSMutableArray alloc] init];
    NSMutableArray *used = [[NSMutableArray alloc] init];

    for (int i = 0; i < chars.count; i++)
    {
        [used addObject:[NSNumber numberWithBool:NO]];
    }

    doPermute(chars, output, used, size, 0);

    return results;
}
-(double) factorialX: (int) num {
    double tempResult = 1;
    for (int i=2; i<=num; i++) {
        tempResult *= i;
    }
    return tempResult;
}
/*=======================Call Like this====================
NSmutableArray=getPermutations("Pass your string", pass your lenght);
NsmutableArray=getPermutations("Janak", 2);
NSMutableArray=[self genValidWords:@"Hello"];
will return all possiable combination of two letters
=======================xxxxxxxxxx====================*/

Answer 3

我得到了解决方案，我想要跟随之类的东西。

-(NSArray*)totalWords:(NSArray*)letters
{
    NSMutableArray *output=[[NSMutableArray alloc]init];

    for (int i=0; i<letters.count; i++)
    {
        for (int j=0; j<letters.count; j++)
        {
            if (i==j) continue;
         for (int k=0; k<letters.count; k++)
         {
             NSString *str=[NSString stringWithFormat:@"%@%@%@",letters[i],letters[j],letters[k]];
             if(i!=j && j!=k && i!=k &&[self checkmeaning:str])
                 [output addObject:str];
            for (int l=0; l<letters.count; l++)
            {
                NSString *str=[NSString stringWithFormat:@"%@%@%@%@",letters[i],letters[j],letters[k],letters[l]];
                if(i!=j && j!=k && i!=k && l!=i && l!=j && l!=k &&[self checkmeaning:str])
                    [output addObject:str];
                for (int m=0; m<letters.count; m++)
                {
                    NSString *str=[NSString stringWithFormat:@"%@%@%@%@%@",letters[i],letters[j],letters[k],letters[l],letters[m]];
                    if(i!=j && j!=k && i!=k && l!=i && l!=j && l!=k && m!=i && m!=j && m!=k && m!=l &&[self checkmeaning:str])
                        [output addObject:str];
                    for (int n=0; n<letters.count; n++)
                    {
                        NSString *str=[NSString stringWithFormat:@"%@%@%@%@%@%@",letters[i],letters[j],letters[k],letters[l],letters[m],letters[n]];
                        if(i!=j && j!=k && i!=k && l!=i && l!=j && l!=k && m!=i && m!=j && m!=k && n!=i && n!=j && n!=k &&n!=m &&n!=l && m!=l&&[self checkmeaning:str])
                        [output addObject:str];
                    }
                }
            }
         }
        }
     }

    NSLog(@"count :%i",[output count]);
    NSLog(@"output array :\n%@",output);
    return output;
}

字母的单词

3 个答案: