假设我有一个像这样的NSNr NSArray：1,2,3
然后所有可能排列的集合看起来像这样：
1,2,3
1,3,2
2,1,3
2,3,1
3,1,2
3,2,1
在objective-c中执行此操作的好方法是什么？
答案 0 :(得分：8)
我使用了Wevah上面的答案中的代码并发现了一些问题，所以我在这里进行了一些修改以使其正常工作：
的NSArray + Permutation.h
@interface NSArray(Permutation)
- (NSArray *)allPermutations;
@end
的NSArray + Permutation.m
#import "NSArray+Permutation.h"
#define MAX_PERMUTATION_COUNT 20000
NSInteger *pc_next_permutation(NSInteger *perm, const NSInteger size);
NSInteger *pc_next_permutation(NSInteger *perm, const NSInteger size)
{
// slide down the array looking for where we're smaller than the next guy
NSInteger pos1;
for (pos1 = size - 1; perm[pos1] >= perm[pos1 + 1] && pos1 > -1; --pos1);
// if this doesn't occur, we've finished our permutations
// the array is reversed: (1, 2, 3, 4) => (4, 3, 2, 1)
if (pos1 == -1)
return NULL;
assert(pos1 >= 0 && pos1 <= size);
NSInteger pos2;
// slide down the array looking for a bigger number than what we found before
for (pos2 = size; perm[pos2] <= perm[pos1] && pos2 > 0; --pos2);
assert(pos2 >= 0 && pos2 <= size);
// swap them
NSInteger tmp = perm[pos1]; perm[pos1] = perm[pos2]; perm[pos2] = tmp;
// now reverse the elements in between by swapping the ends
for (++pos1, pos2 = size; pos1 < pos2; ++pos1, --pos2) {
assert(pos1 >= 0 && pos1 <= size);
assert(pos2 >= 0 && pos2 <= size);
tmp = perm[pos1]; perm[pos1] = perm[pos2]; perm[pos2] = tmp;
}
return perm;
}
@implementation NSArray(Permutation)
- (NSArray *)allPermutations
{
NSInteger size = [self count];
NSInteger *perm = malloc(size * sizeof(NSInteger));
for (NSInteger idx = 0; idx < size; ++idx)
perm[idx] = idx;
NSInteger permutationCount = 0;
--size;
NSMutableArray *perms = [NSMutableArray array];
do {
NSMutableArray *newPerm = [NSMutableArray array];
for (NSInteger i = 0; i <= size; ++i)
[newPerm addObject:[self objectAtIndex:perm[i]]];
[perms addObject:newPerm];
} while ((perm = pc_next_permutation(perm, size)) && ++permutationCount < MAX_PERMUTATION_COUNT);
free(perm);
return perms;
}
@end
答案 1 :(得分：5)
可能有更好的方法（就地或其他），但这似乎有效：
部首：
@interface NSArray (PermutationAdditions)
- (NSArray *)allPermutations;
@end
Implemetation：
@implementation NSArray (PermutationAdditions)
NSInteger *pc_next_permutation(NSInteger *p, const NSInteger size) {
// slide down the array looking for where we're smaller than the next guy
NSInteger i;
for (i = size - 1; p[i] >= p[i + 1]; --i) { }
// if this doesn't occur, we've finished our permutations
// the array is reversed: (1, 2, 3, 4) => (4, 3, 2, 1)
if (i == -1)
return NULL;
NSInteger j;
// slide down the array looking for a bigger number than what we found before
for (j = size; p[j] <= p[i]; --j) { }
// swap them
NSInteger tmp = p[i]; p[i] = p[j]; p[j] = tmp;
// now reverse the elements in between by swapping the ends
for (++i, j = size; i < j; ++i, --j) {
tmp = p[i]; p[i] = p[j]; p[j] = tmp;
}
return p;
}
- (NSArray *)allPermutations {
NSInteger size = [self count];
NSInteger *perm = malloc(size * sizeof(NSInteger));
for (NSInteger idx = 0; idx < size; ++idx)
perm[idx] = idx;
NSInteger j = 0;
--size;
NSMutableArray *perms = [NSMutableArray array];
do {
NSMutableArray *newPerm = [NSMutableArray array];
for (NSInteger i = 0; i <= size; ++i)
[newPerm addObject:[self objectAtIndex:perm[i]]];
[perms addObject:newPerm];
} while ((perm = pc_next_permutation(perm, size)) && ++j);
return perms;
}
@end
要使用它，只需#import标题文件，然后调用[yourArray allPermutations];该方法将返回一个包含每个排列数组的数组。
[代码改编自PHP代码here。]
答案 2 :(得分：4)
您可以将NSString *characters更改为id something并将其用于任何类型的对象
NSArray *array = [NSArray arrayWithObjects:@"a",@"b",@"c", nil];
NSMutableArray *permutations = nil;
int i = 0;
for (i = 0; i < array.count ; i++){
if (!permutations){
permutations = [NSMutableArray array];
for (NSString *character in array){
[permutations addObject:[NSArray arrayWithObject:character]];
}
} else {
//make copy of permutations array and clean og array
NSMutableArray *aCopy = [[permutations copy] autorelease];
[permutations removeAllObjects];
for (NSString *character in array){
//loop through the copy
for (NSArray *oldArray in aCopy){
//check if old string contains looping char..
if ([oldArray containsObject:character] == NO){
//update array
NSMutableArray *newArray = [NSMutableArray arrayWithArray:oldArray];
[newArray addObject:character];
//add to permutations
[permutations addObject:newArray];
}
}
}
}
}
NSLog(@"permutations = \n %@",permutations);
答案 3 :(得分：3)
我最近偶然发现了同样的问题，写了一个我认为更具可读性的递归解决方案。它依赖于注意到的here核心主体。我将它包含在这里以防它可以帮助某人：
- (NSMutableArray <NSArray *> *)permutationOfArray:(NSMutableArray *)array withPrefixArray:(NSMutableArray *)prefixArray withPermutations:(NSMutableArray <NSArray *> *)permutations {
NSUInteger numArrayElements = [array count];
if (numArrayElements == 0) {
[permutations addObject:prefixArray];
return permutations;
}
for (NSUInteger i = 0; i < numArrayElements; i++) {
// Remove object and add it to the prefix array.
// Note: we must create new copies of the arrays as we
// are running using recursive call and these objects
// are called by reference.
NSMutableArray *newArray = [NSMutableArray arrayWithArray:array];
[newArray removeObjectAtIndex:i];
NSMutableArray *newPrefixArray = [NSMutableArray arrayWithArray:prefixArray];
[newPrefixArray addObject:[array objectAtIndex:i]];
[self permutationOfArray:newArray withPrefixArray:newPrefixArray withPermutations:permutations];
}
return permutations;
}
- (NSArray <NSArray *> *)allPermutationsOfArray:(NSArray *)array {
return [self permutationOfArray:[NSMutableArray arrayWithArray:array] withPrefixArray:[NSMutableArray array] withPermutations:[NSMutableArray array]];
}
示例电话：
NSArray <NSArray *> *permutations = [self allPermutationsOfArray:@[@1, @2, @3]];
for (NSArray *singlePermutation in permutations) {
NSLog(@"%@", [singlePermutation componentsJoinedByString:@", "]);
}
结果：
1,2,3
1,3,2
2,1,3
2,3,1
3,1,2
3,2,1
答案 4 :(得分：2)
扩大了SSJ对
的回答 为了清晰起见，-log打印
- 与任何对象一起工作
- 更多解释
- 可能导致错误的边缘情况
[self allPermutationsOfArray:@[@1,@2,@3,@4]]; // usage
...
-(void)allPermutationsOfArray:(NSArray*)array
{
NSMutableArray *permutations = [NSMutableArray new];
for (int i = 0; i < array.count; i++) { // for each item in the array
NSLog(@"iteration %d", i);
if (permutations.count == 0) { // first time only
NSLog(@"creating initial list");
for (id item in array) { // create a 2d array starting with each of the individual items
NSMutableArray* partialList = [NSMutableArray arrayWithObject:item];
[permutations addObject:partialList]; // where array = [1,2,3] permutations = [ [1] [2] [3] ] as a starting point for all options
}
} else { // second and remainder of the loops
NSMutableArray *permutationsCopy = [permutations mutableCopy]; // copy the original array of permutations
[permutations removeAllObjects]; // remove all from original array
for (id item in array) { // for each item in the original list
NSLog(@"adding item %@ where it doesnt exist", item);
for (NSMutableArray *partialList in permutationsCopy) { // loop through the arrays in the copy
if ([partialList containsObject:item] == false) { // add an item to the partial list if its not already
// update a copy of the array
NSMutableArray *newArray = [NSMutableArray arrayWithArray:partialList];
[newArray addObject:item];
// add to the final list of permutations
[permutations addObject:newArray];
}
}
}
}
[self printArrayInLine:permutations];
}
NSLog(@"%lu permutations",(unsigned long)permutations.count);
}
-(void)printArrayInLine:(NSArray*)twoDimensionArray
{
NSString* line = @"\n";
for (NSArray* array in twoDimensionArray) {
line = [line stringByAppendingString:@"["];
for (id item in array) {
line = [line stringByAppendingString:[NSString stringWithFormat:@"%@,",item]];
}
line = [line stringByAppendingString:@"]\n"];
}
NSLog(@"%@", line);
}
输入日志输出@ [@ 1，@ 2，@ 3]
iteration 0
creating initial list
[1,]
[2,]
[3,]
iteration 1
adding item 1 where it doesnt exist and creates a new list
adding item 2 where it doesnt exist and creates a new list
adding item 3 where it doesnt exist and creates a new list
[2,1,]
[3,1,]
[1,2,]
[3,2,]
[1,3,]
[2,3,]
iteration 2
adding item 1 where it doesnt exist and creates a new list
adding item 2 where it doesnt exist and creates a new list
adding item 3 where it doesnt exist and creates a new list
[3,2,1,]
[2,3,1,]
[3,1,2,]
[1,3,2,]
[2,1,3,]
[1,2,3,]
6 permutations