Objective-C的拆分阵列分成两个独立的阵列基于奇/偶指标 [英] Objective-C Split an array into two separate arrays based on even/odd indexes

问题描述

我有一个数组，NSMutableArray的*字符串数组，看起来像这样

I have an array, NSMutableArray *stringArray that looks like this

stringArray = 

[0]String1
[1]String2
[2]String3
[3]String4
[4]String5
[5]String6

这个数组转换成两个数组的基础上，即使我将如何着手拆分/奇索引？

How would I go about splitting this array into two arrays based on even/odd indexes?

例如：

NSMutableArray *oddArray = ([1], [3], [5]);

NSMutableArray *evenArray = ([0], [2], [4]);

在此先感谢！

推荐答案

有以下方法可以做到这一点： -

There are following ways you can achieve that:-

<强>第一和第二溶液都已经由上述两个提及。下面是相同的实施： -

//First Solution
NSArray *ar=@[@"1",@"2",@"3",@"4",@"5"];
NSMutableArray *mut1=[NSMutableArray array];
NSMutableArray *mut2=[NSMutableArray array];
[ar enumerateObjectsUsingBlock:^(id object, NSUInteger idx, BOOL *stop) {
    if (idx%2==0)
    {
        [mut1 addObject:object];
    }
    else
    {
        [mut2 addObject:object];
    }
}];

//Second Solution
NSMutableIndexSet *idx1 = [NSMutableIndexSet indexSet];
NSMutableIndexSet *idx2 = [NSMutableIndexSet indexSet];
for (NSUInteger index=0; index <ar.count(); index++)
{
   if(index%2==0)
    {
        [idx1 addIndex:index];
    }
    else{
        [idx2 addIndex:index];
    }
}
NSArray *evenArr=[ar objectsAtIndexes:idx1];
NSArray *oddArr=[ar objectsAtIndexes:idx2];
NSLog(@"%@",evenArr);
NSLog(@"%@",oddArr);

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