排序数组字典

本文介绍了排序数组字典的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！ 问题描述

我有许多字符串数组。我wan't将它们整理成一本字典，所以所有开始同一封信去到一个数组，然后数组成为一个关键的字符串值;关键是与其中的所有词的价值的数组开头字母。

示例

键=A＆GT;＆GT;值=数组=苹果，动物，字母，ABC ......答案=B＆GT;＆GT;值=数组=蝙蝠，球，香蕉......

我怎样才能做到这一点？非常感谢事先！

解决方案

的NSArray *名单= [NSArray的arrayWithObjects：@苹果，动物，蝙蝠，球，零]*的NSMutableDictionary字典=的NSMutableDictionary字典]对于（的NSString *单词）{    的NSString * firstLetter = [[字substringToIndex：1] uppercaseString];    NSMutableArray里* letterList = [字典objectForKey：firstLetter];    如果（！letterList）{        letterList = [NSMutableArray的阵列]        [字典的setObject：letterList forKey：firstLetter];    }    [letterList ADDOBJECT：字]。}的NSLog（@％@，字典）;

I have and array of many strings. I wan't to sort them into a dictionary, so all strings starting the same letter go into one array and then the array becomes the value for a key; the key would be the letter with which all the words in it's value's array begin.

Example

Key = "A" >> Value = "array = apple, animal, alphabet, abc ..." Key = "B" >> Value = "array = bat, ball, banana ..."

How can I do that? Thanks a lot in advance!

解决方案

NSArray *list = [NSArray arrayWithObjects:@"apple, animal, bat, ball", nil]; NSMutableDictionary *dict = [NSMutableDictionary dictionary]; for (NSString *word in list) { NSString *firstLetter = [[word substringToIndex:1] uppercaseString]; NSMutableArray *letterList = [dict objectForKey:firstLetter]; if (!letterList) { letterList = [NSMutableArray array]; [dict setObject:letterList forKey:firstLetter]; } [letterList addObject:word]; } NSLog(@"%@", dict);