iOS开发-面试总结(十四)

面试总结(四)实现有序字典

实现方法

  1. 按NSDictionary的key来对其进行排序:
  2. 将字典的Value再放到一个字典里面,key分别使用有序的字符串
  3. 先将字典转模型,再放到数组里面
  4. 先将dict的allkeys赋给一个数组,然后通过sortedArrayUsingComparator:方法对数组排序,然后遍历数组取字典对应key的值就ok


这里只说大概的两张方法的实现

方法一

复制代码
1 NSArray *keys = [dict allKeys];
2 NSArray *sortedArray = [keys sortedArrayUsingComparator:^NSComparisonResult(idobj1, id obj2) {
3 return [obj1 compare:obj2 options:NSNumericSearch];
4 }];
5 for (NSString *categoryId in sortedArray) {
6 ……
7 NSLog(@"[dict objectForKey:categoryId] === %@",[dictobjectForKey:categoryId]);
8 }
复制代码


方法二:

1 NSString *filePath = [[NSBundle mainBundle]pathForResource:@"students" ofType:@"plist"];
2 NSDictionary *stuDic = [NSDictionary dictionaryWithContentsOfFile:filePath];
3 self.studentIndexArray = [NSMutableArray arrayWithCapacity:1];
4 NSArray *arr = [[stuDic allKeys] sortedArrayUsingSelector:@selector(compare:)];
原文地址:https://www.cnblogs.com/stronger-ios-lcx/p/5634959.html